Lab Report Density Lab

Lab Report Density Lab
The purpose of the Density/Graphing Mini Lab was to determine how the density of a small amount
of a substance relates to the density of a large amount of the same substance. The density of a
substance is determined by the mass (resistance to change in acceleration) and the volume (how
much space it takes up). To calculate density the formula used is, F = m/v. Mass is measured in g,
volume is measured in mL, and density is measured in cm³. The hypothesis, "If the density of a
small amount of a substance is compared to the density of a large amount of the same substance then
they densities will be the same because the ratio between mass and volume remains the same," was
proven true as when mass is increased, volume increased as well at the same ... Show more content
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Marble 5; 5.1g; total mass is 25.9g; total volume is 10.5mL, and volume of Marble 5 is 2 mL.
Marble 6; 5.6g; total mass is 31.5g; total volume is 13mL, and volume of Marble 6 is 2.5mL. When
analyzing the results it was discovered that as the number of marbles increase the total volume
increased due to the increase of particles. The graph shows a straight line which also demonstrates
how the ratio of mass and volume will not cause a change in the density no matter how much of the
substance is being used. Possible errors that could have skewed the data include the water not being
30mL at first or a broken calculator. If the water was not originally 30 mL then the total volume of
the marbles would be incorrect as the number wouldn't be accurate because of uneven starting
points. If the calculator used to find the answers is broken then the answers would all be inaccurate,
which would mess up the data. Possible ways to fix these errors could be making sure that when
reading the graduated cylinder, it is read at eye level and the bend, or meniscus, is what is read. If
the highest point is read the data would be incorrect because the water is holding on to the sides of
the
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Particles Lab
Pressure vs. Number of Particles Lab Report
Introduction: The experiment deals with many different parts and one of them is gas, a substance
that moves freel through the air and their particles are stread farther away from each other when
compared to the other states of matter. The difference between the states of matter are how close the
particles are to each other. The experiment was completed to find how pressure affects the number
of particles.
Pressure, which is the force exerted on an object by physical contact, was used to apply force into
the pressure sensor. A pressure sensor measures the amount of force can object has and this will tell
us how the pressure is affecting the number of particles. Also the volume, of the pressure ... Show
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When the pressure was added to the pressure sensor, it increased the particles speed and added more
particles too. Also the object that withheld the pressure had a constant volume, meaning the shape
did not change Then when the student increased the pressure the greater number of particles. When
the air temperature increases, the particles sped up and this created greater force during the
collisions. Also the graph was supposed to start at zero however the y–intercept was off by a little
because of an error that occurred in the lab. But overall, the graph expressed what was expected and
shows a linear trendline that

Surface Area to Volume Ratio and the Relation to the Rate...
Surface Area to Volume Ratio and the Relation to the Rate of Diffusion
Aim and Background
This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion
and how this relates to the size and shape of living organisms.
The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to
diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter
because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to
diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for
the nutrients and oxygen to diffuse into the cell so the cell would probably not ... Show more content
The larger blocks have a smaller surface area than the smaller blocks. The smallest block has 1.2mm
squared of surface area for every 1mm cubed of volume. The largest block only has 0.2mm squared
of surface area for each 1mm cubed of volume. This means that the hydrochloric acid is able to
diffuse the smallest block much faster than the largest block.
When the Surface Area/Volume Ratio goes down it takes longer for the

Lab Report Discussion Essay
The purpose of this lab was to create an experiment to test samples of plastic–like cylinders, to see if
the different masses had an effect on the density of the objects. The original hypothesis was if the
mass of the substance increased there would be no effect on its density.
As seen in the data analysis, the average density was 1.7 (g/mL), with only a negligible 0.1 (g/mL)
difference across all four masses of the substance. Furthermore in the data, if the mass of the
substance increased, the volume increased proportionally with it. This information ultimately
supports our hypothesis. As addressed during classroom lectures, the relationship between the two
variables is most likely explained by the fact that density is a ratio. It would be logical to assume
that while the mass and volume increase at the same rate, the division of mass by volume would
result in nearly the same number. ... Show more content on Helpwriting.net ...
However, while the accuracy of the calculated average density is unknown, it was very precise,
given the negligible 0.1 (g/mL) inconsistency. Possible sources of errors could result from
inaccurate following of the procedure, or a lack of good time management. These would result in
human errors such as rushing and taking inaccurate measurements, both which would result in
flawed raw data. This could be remedied by carefully following each step in the procedure and
slowing down movements to reduce measurement error. However, do not believe we had many
flaws in the execution of our

Research on Density Essay
Density: Using Experimental Techniques to Solve an Inquiry based problem
ABSTRACT
The topic of this experiment is Density. The objective is to find two ways in which the density of a
given object can be determined, and to find out which of the two ways is more accurate and hence
better to use in such a case. The two methods used in this experiment are finding the dimensions of
the object and water displacement. These are two ways of finding the volume of an object, and they
were chosen since the density of an object may be found using its mass and its volume. The
experiment yielded two different density values, however when error analysis was conducted, the
water displacement method was proven to be more accurate.
INTRODUCTION ... Show more content on Helpwriting.net ...
Repeat measuring the mass of the object three (3) more times; ensuring that the balance is re–zeroed
after each measurement. Tabulate data gathered. Procedure 2: Find the mass of the object given, as
done in procedure 1. Ensure that the object is dry when it is being placed on the balance, as this will
yield an inaccurate reading. Get a clean 1000milliter beaker. Fill the beaker to approximately half its
capacity. Note the exact volume of water placed into the beaker. When reading the volume of the
water, ensure that this is done at eye level, and that it is read at the bottom of the meniscus. After
noting the initial volume of water, place the object carefully into the water. DO NOT SPLASH. Note
the final volume of water in the beaker. Tabulate data gathered.
DATA/RESULTS
Table 1: Mass of Given Object for four(4) Separate Trials TRIAL NUMBER 1 2 3 4 MASS(g) 5.32
5.32 5.32 5.32
Table 2: Measurements of Dimensions of the Given Object DIMENSION Length– b Width– w
Height– h Table 3: Initial and Final Volumes of Water INTIAL(mL) 500.0 Calculations: Volume1 =
area of face of solid x width = (1/2b x h) x w = (1/2 x 10.0cm x5.0cm) x 3.5 = 87.50 cm3 Volume2 =
Vf – Vi = 520.0mL– 500.0mL = 20.0mL Density = Mass/Volume D = m/V D1 = m/V1 ; m=5.32g,
V1=87.50cm3 FINAL(mL) 520.0 MEASUREMENT(cm) 10.0 3.5 5.0
Therefore D1= 5.32g 87.50cm3 = 0.0608 g/cm3 D2 = m/V2 ; m= 5.32g, V2= 20.00 mL Therefore
D2 = 5.32g 20.00mL =0.266g/mL
Error

Effect of Surface Area to Volume Ratio on Rate of Osmosis
[Type text] [Type text] [Type text]
_An experiment on the effect of surface area to volume ratio on the rate of osmosis of Solanum
tuberosum L._
BACKGROUND
A cell needs to perform diffusion in order to survive. Substances, including water, ions, and
molecules that are required for cellular activities, can enter and leave cells by a passive process such
as diffusion. Diffusion is random movement of molecules in a net direction from a region of higher
concentration to a region of lower concentration order to reach equilibrium. Diffusion does not
require any energy input. Diffusion is needed for basic cell functions – for example, in humans, cells
obtain oxygen via diffusion from the alveoli of the lungs into the blood and in plants water ... Show
OBSERVATIONS
There were no qualitative changes to the potato that could be detected by the five senses after the
cubes were taken out of the water
The water level in the beakers did not have any significant decrease
RESULTS (PROCESSED DATA)
Cube Size (cm)
Surface Area (cm2)
Volume (cm3)
Surface Area : Volume
Mass Before (g)
Mass After (g)

Percentage Change (%)
1 x 1 x 1
6
1
6:1
0.96
1.07
11
1.5 x 1.5 x 1.5
13.5
3.375
4:1
3.47
3.62
4.27
2 x 2 x 2
24
8
3:1
7.76
8.26
7.66
2.5 x 2.5 x 2.5

37.5
15.625
2.4:1
14.22
14.93
5.02
3 x 3 x 3
54
27
2:1
25.74
27.05
5.11
Percentage Change calculation:
Average Percentage Change:
CONCLUSION
In conclusion, the potato cube with the highest surface area to volume ratio (the 1x1x1 cube) had the
fastest rate of diffusion as it had the largest percentage increase in mass. While all the other cubes of
potato had larger increases in mass at face value compared to the smallest cube, the smallest cube
had the largest overall gain in percentage. The results support my hypothesis that the smallest cube
will have a higher rate of osmosis because it has a proportionally larger amount of surface area
compared to its volume.
As seen from the graph above, despite my hypothesis being correct, the trend in the percentage
increase of mass was not a steady

Chemistry Experiment Essay
Lab Report Experiment 1 & 2 CHM–101 Amele Takpara Partner: Jessamyn Dupree
––––––––––––––––––––––––––––––––––––––––––––––––– Experiment 1
––––––––––––––––––––––––––––––––––––––––––––––––– (Pre–Lab Questions) 1. In the design
of a Bunsen burner, explain the purpose of a. the gas control valve The gas control valve regulates
the rate at which methane enters the burner. b. and the air vents. The air vents control the rate at
which air enters the burner 2. Why is a luminous yellow flame often ''smoky''? The luminous yellow
flame is smoky because no air is entering the burner and hydrocarbon is converted into carbon
dioxide 3. A student wanted 20.000 g of a salt. Which balance should the student use in order to ...
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The difference between an intensive property and an extensive property is that the extensive
property of a substance whereas an intensive property does not. Density is an example of intensive
property of a substance Extensive property example: mass 2. In order to calculate the density of a
solid or liquid sample, what measurements are needed? Mass and volume 3. The volume of a fixed
mass of a liquid sample increases as the temperature rises from 20 to 408C. Does the density
increase, decrease, or stay the same? Explain your answer. The density decreases. Density of the
liquid sample is the mass divide by the volume. If the mass stays constant and the volume increases
then the density of the liquid sample will decrease. 4. A solid block of exactly 100.0 cm3 has a mass
of 153.6 g. Determine its density. Will the block sink or float on water? Density (d)=m/v = 153.6g /
100cm3 = 1.536 g/cm3 Since this solid block is more dense than water (1.536 g/cm3 > 1 g/cm3)
it will sink. 5. A salvage operator recovered coins believed to be gold. A sample weighed 129.6 g
and had a volume of 15.3 cm3. Were the coins gold (d = 19.3 g/cm3) or just yellow brass (d = 8.47
g/cm3)? Show your work. Density (d)=m/v = 129.6g / 15.3cm3 = 8.47 g/cm3 The density of the
coins is 8.47 g/cm3 which matches the density of yellow brass the coins recovered are yellow brass
The purpose of the

Density Lab Report Essay
Abstract
In Measuring and Understanding Density, several experiments were performed to find density of
regularly shaped objects, irregularly shaped objects, liquids and gasses. An additional experiment
was done to find the specific gravity of a sampling of liquids. The purpose of the experiment was to
provide a better understanding of density and to be able to extrapolate unknowns based upon these
calculations. The experiments yielded data in keeping with Kinetic–molecular theory in regards to
the density of water versus its temperature. Key measurements and formulae were also used to
determine densities of metal and plastic objects as well as irregularly shaped rocks. It is possible to
find the density of an object (be it liquid, gas or ... Show more content on Helpwriting.net ...
With volume and mass determined, the density formula (d = m/V) was used to determine the
densities of each rock and recorded in g/ml. In the third stage of this experiment, the density of a
liquid was determined and compared to known standards. A 100ml beaker was filled to about half–
full with room–temperature distilled water. The temperature of the water in ◦C was recorded in order
to compare to known standards later. A 50ml beaker was then weighed on a scale in order to
determine mass and recorded. A sample of the distilled water with an exact volume of 10ml was
then placed in the 50ml beaker using a volumetric pipette. The 50ml beaker with the 10ml of water
was then weighed again and the initial mass of the beaker was subtracted from this mass to obtain
the mass of the 10ml of water. With the volume and the mass of the water now known, density was
calculated using d = m/V and recorded in g/ml. This process was then repeated to check for
precision and compared to standard values to check for accuracy. Standard values were obtained
from CRC Handbook, 88th Ed. In the fourth stage of this experiment, the density of a gas was
determined. A 250ml flask was weighed with an empty rubber balloon and the mass was recorded.

Surface Area To Volume Ratio
Single celled organisms are small meaning that their surface area is big compared to their volume.
These organisms have a large surface area to volume ratio which means that they can obtain
different substances by diffusion through their relatively large plasma membrane. The substances
have to diffuse only short distances so they can diffuse at a faster rate and meet the organism's
needs. Multi–cellular organisms have a much smaller surface area to volume ratio. Many of their
cells are not in direct contact with their surroundings so they cannot solely rely on diffusion to
supply all of their organs with oxygen and nutrients, as the distance from their surface to all cells is
too far. Humans are multi–cellular and have special surfaces for ... Show more content on
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– A mechanism to maintain mass flow in one direction
– A closed system of tubular vessels with a branched network allowing supply to all areas of
organism
– A means of controlling the flow to suit the needs of the organism
What is blood pressure?
Blood pressure is a measure of the pressure exerted by the blood against artery walls. Blood varies
with heartbeat strength, age, blood volume, health and fitness. Blood pressure is measured as two
numbers, e.g. 120/80 these numbers represent systolic and diastolic the big number is systolic which
the blood pressure is during a heartbeat and the smaller number is diastolic which the blood pressure
in–between heartbeats is. The numbers show pressure reading in mm of mercury.
How does blood pressure affect health?
High blood pressure can make blood vessels more likely to burst, and can cause strokes and kidney
damage. Low blood pressure can cause dizziness, fainting and poor circulation.
Capillaries
These are found all over the body and are essential for the exchange of materials between the blood
and other body cells.
Capillaries are so small that they can only be seen using a

Mathematical Modeling Of A Functional Building
Mathematical Modeling of a Functional Building
Reminder Abstract should be able to be taken out of the paper and the rest of the paper still makes
sense.
Throughout architecture there is a delicate balance between functionality and aesthetics. However a
rule of thumb to follow is "form follows function". Therefore for the purpose of our designed office
block, we decided to allow form to follow function for majority of our decisions, but we also kept
the aesthetics in our modeling conscious.
Abstract and introduction: what a reader would read for the goal of the paper. We did this and then
this and then this... Can divulge all the aspects of the problem, but not too much detail.
Introduction:
We are two partners in an independent ... Show more content on Helpwriting.net ...
We finally settled on the semicircle equation because changing the height of the structure without
altering the set width, allowed for a more natural looking curve, whereas the function creates a more
"pinched' look. (include pictures of the buildings)
36* sin( π/ 72* x)
46* sin( π/ 72* x)
51* sin( π/ 72* x)
Ultimately, we made this choice for the purposes of this structure because it not only provides an
ideal form, but also allows for diverse functions as well.
Because we chose this graph to model the main arch of the building, we wanted to experiment in a
sense, and see how we could use this model in conjunction with the rectangular base to adhere to
potentially different functions. One of those possibilities was for the "warehouse" model. To start off
we just derived our original function in order to find the maximum x value, and its corresponding y
coordinate to ultimately find the volume of the warehouse at that given height of the model. The
derivative function of the semicircle for any height (greater than or equal to 36, less than or equal to
54) is shown below:
We used this equation to generate different heights that would maximize the area of the cuboid. We
were able to isolate just the area of the cuboid because that one face is simply projected backwards
150m, so we mathematically we were able to ignore that constant in the volume equation shown
above. With the help of Excel spreadsheets, we calculated the

Essay about Lab 1-Measurement, Length, Mass, Volume,...
GS104
Lab Report
Experiment # 1
Measurement, Length, Mass, Volume,
Density & Time
David Case:
January 17, 2015
Experiment #1
Measurement, Length, Mass, Volume, Density, & Time
Objectives:
To make basic distance, mass, density, and time measurements.
To make calculations of volume and density, using proper units, and to practice using graphing
software while graphing the relationship between the circumference of a circle and its diameter.
Materials: stopwatch meter tape 3 rectangular objects metric ruler pencils metal bolt 500‐G scale
graduated cylinder five circular objects
Lab Notes: 1. Estimation of Various ... Show more content on Helpwriting.net ...
By having the length, width, and height measurements off by a little, the volume would be off by a
huge amount. A small error in the hand measurements would grow by leaps and bounds in
measuring volume. 3. Graphing data and the determination of π In this experiment, I measured 5
round objects of varying sizes. The objects I selected are a cup, a bowl, a soda can, a babies rattle, a
CD.
Object
Diameter (cm)
Circumference (cm)
Cup
7
22
Bowl
14
48
Can

6
21
Rattle
5
18
CD
12
38
Question: What is the slope of "C vs d" line? What does it represent? The slope of the "C vs d" line,
according to the above graph, is 3.2548. This represents a ratio between the circumference and the
diameter of a round object. Being that I used imprecise measuring devices my measurements are a
little off. Although my ratio seems consistent in my graph. With a more accurate measuring device
my measurement would be closer to π.
G. Percentage error formula: Percentage error = π = 3.14 Estimate = 3.2548 3.14–3.2548/3.14 = ‐
0.0365 x ‐1 = 0.0365 x 100% = 3.65% My percent of error is 3.65%. 4 Density Measurements:
determine the density of a metal bolt by water‐displacement method:
A, B & C: For this experiment first I weighed a metal bolt using a digital scale, getting a weight of
7.8 g. Then I filled the graduated cylinder with water to 10–mL. Next I placed the metal bolt into the
cylinder and recorded the new volume. This reading was 12–mL.

Surface Area Volume Ratio Experiment
Surface area / Volume ratio Experiment
Introduction:
The surface area to volume ratio in living organisms is very important. Nutrients and oxygen need to
diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter
because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to
diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for
the nutrients and oxygen to diffuse into the cell so the cell would probably not survive. Single celled
organisms can survive as they have a large enough surface area to allow all the oxygen and nutrients
they need to diffuse through. Larger multi celled organisms need specialist ... Show more content on
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Interpretation:
In all the blocks of gelatin the rate of penetration of the hydrochloric acid from each side would
have been the same but all the blocks take different amounts of time to clear because they are
different sizes. As the blocks get bigger it takes longer for the hydrochloric acid to diffuse through
the entire block and so clear the dye. It takes longer to reach the centre of the cube even though the
rate of diffusion is the same for all the cubes.
As the volume of the blocks goes up the Surface Area / Volume ratio goes down. The larger blocks
have a smaller proportion of surface area than the smaller blocks.
When the surface area to volume ratio goes down it takes longer for the hydrochloric acid to diffuse
into the cube but if the ratio goes up then the hydrochloric acid diffuses more quickly into the block
of gelatin. Some shapes have a larger surface area to volume ratio so the shape of the object can
have an effect on the rate of diffusion.
By increasing the surface area the rate of diffusion will go up.
Precautions:
All the

Essay on Boyles Law Apparatus
Abstract The objective of this lab was to determine the relationship (if any), between the pressure
and volume of a gas given the temperature and # of molecules remained constant. Using the Boyle's
law apparatus, and textbooks to demonstrate pressure it was concluded that there was a relationship
between pressure and volume. However, the relationship was not a direct relationship, and it was
determined that the pressure and volume of a gas are inversely proportioned. Thus,proving Boyle's
theory correct. Introduction
Objectives: The main objective of this lab was to determine the relationship between the volume and
pressure when the temperature and number of molecules remains the same throughout. Other minor
objectives of ... Show more content on Helpwriting.net ...
Don't insert the yellow stopper so far that the wooden block can't rest stably on the lab bench. Place
the smaller wooden block on top of the apparatus.
1. 2 – Remove the pin cover from the graduated cylinder and press down or pull up until the
plunger reaches 30mL (cm3). Replace the pin TIGHTLY. Test the plunger by pressing down on the
small wooden block to make sure that when you let go, it always returns to about 30 mL.
(29 ½ is ok and some experimental error. ) This is volume at 0 textbooks (tbs) of pressure. Add 2.4
tbs of pressure to account for the atmosphere and record your data.
3 – Add one textbook to the center of the small wood platform and record the new volume of the air
that is occupying the cylinder. Record the pressure always adding 2.4 tbs to how many tbs used to
account for the atmosphere. If necessary, gently hold the book in place.
4 – Repeat step 3 except now with 2 textbooks, and then 3 textbooks, and finally 4 textbooks.
5 – Start the experiment over from step 2 so that you can obtain more trials. Do the experiment a
total of 4 times for each textbook load. Average your results for each textbook load.
Results
Table #1: Trial #1 Trial #2 Trial #3 Trial #4 Trial #5 Average
2.4tbs 29mL 29mL 29mL 29mL 29mL 29mL
3.4tbs 23.5mL 23.5mL 23.0mL 24.0mL 23.0mL 23.4mL
Table #2:

Maths Test Papers Ix(9) Std.
Solved Paper−1 Class 9th, Mathematics, SA−2
Time: 3hours Max. Marks 90 General Instructions 1. All questions are compulsory. 2. Draw neat
labeled diagram wherever necessary to explain your answer. 3. Q.No. 1 to 8 are of objective type
questions, carrying 1 mark each. 4. Q.No.9 to 14 are of short answer type questions, carrying 2
marks each. 5. Q. No. 15 to 24 carry 3 marks each. Q. No. 25 to 34 carry 4 marks each.
1.
Point (–3, 5) lies in the (A) first quadrant (C) third quadrant
(B) second quadrant (D) fourth quadrant
2.
If ∆ ABC ≅ ∆ PQR and ∆ ABC is not congruent to ∆ RPQ, then which of the following is not true:
(A) BC = PQ (B) AC = PR (C) QR = BC (D) AB = PQ ABCD is a cyclic quadrilateral such that AB
is a diameter of the circle ... Show more content on Helpwriting.net ...
(ii) In the classroom discuss what conclusions can be arrived at from the graph. 20. The taxi fare in a
city is as follows: For the first kilometre, the fares is Rs 8 and for the subsequent distance it is Rs 5
per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this
information, and draw its graph. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m 22.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the
given figure). Show that:
21.
(i) ∆APD ≅ ∆CQB (ii) AP = CQ (iii) ∆AQB ≅ ∆CPD (iv) AQ = CP (v) APCQ is a parallelogram
23. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal
BD (See the given figure). Show that
(i) ∆APB ≅ ∆CQD
Material downloaded from http://myCBSEguide.com and http://onlineteachers.co.in Portal for
CBSE Notes, Test Papers, Sample Papers, Tips and Tricks
(ii) AP = CQ 24. The following table gives the distribution of students of two sections according

to the mark obtained by them: Section A Marks 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 Frequency 3
9 17 12 9 Section B Marks 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 Frequency 5 19 15 10 1
Represent the marks of the students of both

The Relationship Between Density And Sugar Content
The Relationship Between Density and Sugar Content
Danielle Shaub
Partners: Emily Alexander, Bill Dempsey, Shayla Ho
Lab Performed 03/11/15 & Lab Report 04/07/15
Abstract:
Using solutions of sugar and water alongside cola, density and percent sugar can be related. The
purpose of this lab is to discover how and with what benefit these factors can be related. Using
assorted measurements and the standard curve of the calculated densities and percentages of sugar
solutions, it was learned that cola has an approximate percent sugar of four percent, according to the
calculations of the lab. Other results included actual percent sugar of soda and various statistics on
measured sugar solutions. Introduction
The purpose of this lab is to explore how density can be used to measure sugar content and percent
sugar of beverages not only in the lab, but in the "real world." Using the standard curve of the
percent sugar and density of the solutions in the lab, the percent sugar of regular Coke was
determined. The standard curve represented the general trend in the correlation of the values
calculated in the lab. Using the data from the experiment, the actual amount of sugar in the soda
could be calculated by using the percent sugar found by using the standard curve. (If the soda has a
mass of x and the percent sugar is the value y, x ● y = mass of sugar.)
Of course, the simple calculation of percent sugar dividing the mass of the sugar by the mass of the
solution can be

How Does the Surface Area to Volume Ratio Affect Heat Loss...
Biology Lab Report BY: Michael Ryan Pranata 11C | Background: As heat is a form of thermal
energy, they tend to have the behavior of reaching a thermal equilibrium. This means that when two
bodies of different temperatures come in contact with each other, the hotter ones will transfer heat
particles to the body with a colder temperature, with an aim to reach this "thermal equilibrium",
whatever the temperature may be. The larger the surface area, means there can be more "paths" from
the sides of the body that are capable of releasing this heat particles, and reaching thermal
equilibrium faster. This is what happens when a hotter body is subjected to a colder one. Research
Question: How does the surface area to volume ratio affect ... Show more content on
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The trial with less time would have a lower rate of heat loss, and vice versa. Materials Required: 1.
Four round bottom flasks of different sizes 2. Hot water 3. Cotton 4. Rubber band 5. Metal stand 6.
Wooden block Procedures to be followed: 1. Prepare all the apparatus required on the table before
starting the experiment 2. Heat the water until it boils 3. While water is heating, start covering the 4
different round bottom flasks with cotton 4. Secure them with a rubber band 5. Fix the retort stand
by putting 2 clamps, one for the flask, and one for the temperature sensor 6. Start preparing the data
logger. The settings in the data logger should be 10 seconds per sample, or 0.1 per second. The time
for one trial should be exactly 180 seconds. 7. Starting with the 50 ml flask, secure this flask to one
of the clamps. 8. Pour an appropriate amount of hot water according to the size of the round bottom
flask 9. Secure the temperature sensor from the data logger to the other clamp. 10. Place the
temperature sensor into the hot water. Make sure the temperature sensor enters the water, not
touching the bottom, or any of the sides of the flask. 11. Start the data logger and wait till the initial
temperature is 77⁰C. 12. Once the temperature of the water is 77⁰C, start recording the data in the
data logger and save it in a file. 13. If the initial temperature is lower or higher than 77 (a margin of

Laboratory Techniques & Measurements Essay
Title: Laboratory Techniques & Measurements
Purpose: To become familiar with the International System of Units and common laboratory
equipment and techniques. To learn how to determine volume, mass, length, and temperature of a
wide variety of items. To learn how to calculate density and concentration of dilutions.
Procedure: I used a ruler, thermometer, and scale to take measurements. I used a graduated cylinder,
short step pipet, scale, and ruler to determine volume and density. I used a volumetric flask,
graduated pipet, pipet bulb, scale, and glass beaker to determine concentrations and densities of
various dilutions.
Data Tables:
|Data Table 1: Length ... Show more content on Helpwriting.net ...
|
|Ice water – 1 minute |5°C |41°F |278°K |
|Ice water – 5 minutes |5.5°C |41.9°F |278.5°K |
|Data Table 3: Mass |
|Object |Estimated Mass |Actual Mass |Actual Mass |
|Pen or Pencil |7g |12.2g |0.0122kg |
|3 Pennies |12g |8.1g |0.0081kg |
|1 Quarter |15g |5.7g |0.0057kg |
|2 Quarters, 3 Dimes |25g |18.2g |0.0182kg |
|4 Dimes, 5 Pennies |20g |22.2g |0.0222kg |
|3 Quarters, 1 Dime, 5 Pennies |65g |32.4g |0.0324kg |

The Density Of Water Is Its Mass Divided By Its Volume
The density of water is its mass divided by its volume. In this case, the mass of water was
determined by subtracting the final mass of the glassware with water by the initial mass of the
glassware without water. The volume was determined by observing closely which line (marked on
the sides of each glassware) the water reached and reporting the most precise measurement. Based
on the class data, the average densities of water and its standard deviations were: 50–mL beaker –
avg: 0.90 g/mL, stand dev: 0.12, 10–mL graduated cylinder – avg: 0.980 g/mL, stand dev: 0.0600,
10–mL volumetric pipette– avg: 0.9800 g/mL, stand. dev: 0.06499, 50–mL burette– avg: 0.969
g/mL, stand. dev: 0.140. The average density of water compares our experimental value to the true
value. This means that the closer our values are to the true density of water, which is approximately
0.998 g/mL at 18.89°C, the more accurate our data is. Furthermore, the standard deviation measures
how precise our results are with one another. Based on the four data sets of the class, the most
precise volumetric glassware is the 10–mL graduated cylinder because its standard deviation is the
lowest. This lower value indicates that this glassware consistently yielded similar densities of water
more often than other glassware. Although tied with the volumetric pipette, the graduated cylinder is
also the most accurate volumetric glassware because the resulting average of its water densities is
closest to the real density of

Flexibility Quantification : Measure By Type Vs. Holistic...
2.3.2 Flexibility quantification: measure by type vs. holistic measure
2.3.2.1 Flexibility measurement: measure by type
1. Volume flexibility
Descriptive measure of volume flexibility can be the lower and upper bound of capacity for range
dimension and time/cost of capacity adjustment for response perspective. But economic metrics are
most commonly used in literature, since many researchers define volume flexibility as the ability to
profitably produce at different levels. Stigler(1939) considered a plant to be flexible if it has a
relatively flat average cost curve. Falkner (1986) suggest the stability of manufacturing costs over
widely varying levels of total production volume to be the measure of volume flexibility. Sethi and
Sethi (1990) developed volume flexibility metric to be the curvature of the average cost curve.
There're also metrics that consider the interaction with demand market. Marschak and Nelson (1962)
argued the value of volume flexibility increases as the variation in market price increases and as the
ability to predict market price before making an output decision increases. Gerwin (1987) measures
it by the ratio of average volume fluctuations over a given period of time to the production capacity
limit.
2. Mix flexibility
There exists a lot of measures on mix flexibility range, the number of product types a system can
produce. (Ettlie 1988, Jaikumar 1986), the size of the universe of products the system is capable of
producing (e.g. Chatterjee et

Artificial Sweetener Vs Coke Essay
Results and Discussion
In this experiment, aspects of the scientific method were introduced and the significance of the
process of gathering measurements during an experiment were illustrated through the use of a
diverse set of glassware: graduated cylinder, volumetric pipette, and burette. These unique
measuring instruments played a role in concluding the densities of the two soft drinks, Coke and
Diet Coke, and whether or not a statistical difference existed between the two substances.
Conversely, the resulting densities determined by each of the differing glassware were assessed and
compared to each other in terms of accuracy and precision, with the most precise tool that produced
consistent similar results, deduced.
With regards to the ... Show more content on Helpwriting.net ...
However, what was not mentioned was how the ingredients specifically contribute.
When analyzing the key ingredients of Coke and Diet Coke, sugar and artificial sweetener,
respectively, equal amounts of each product produces different levels of sweetness, with artificial
sweeteners being over a hundred times sweeter than natural sugars.1 As a result, in order to produce
a similar level of taste between the two soft drinks, a large amount of sugar is added into Coke, in
comparison to no sugar, and a small amount of sweetener in Diet Coke. Thus contributing to a
greater mass of Coke, and therefore a greater density, despite the equal volumes of both soft drinks.
Consequently, if the volume increased by a doubled amount under similar experimental conditions
where temperature and pressure remained constant, the density of the soda would not be expected to
change due to density being an intensive property that is a ratio between the mass over the volume
of a substance.2 As an intensive property, density, under similar experimental conditions, will be the
same without regards to the amount presented. Doubling the volume would only serve to punctuate
the concept, considering how dividing two extensive properties, mass and volume, that are directly
proportional to the quantity of a substance, will cancel out the quantity dependence, resulting in a
ratio that is similar each

Labratory Techniques and Measurments
Jessica Loper Date of Experiment: February 5, 2013
Report Submitted: February 11, 2013
Title: Laboratory Techniques and Measurements
Purpose: To gain knowledge about the International System of Units and use it to determine volume,
mass, length, and temperature. To learn to use these forms of measurement to determine density and
concentration, as well as learning basic lab equipment to create dilutions.
Procedure:
Part 1: After reading the various information given, I gathered different objects and measured them
using both centimeters and millimeters then converted them to meters. Next, I recorded the
temperature of hot tap water and boiling water as ... Show more content on Helpwriting.net ...
The water could resist boiling at 100°C because you could be above or below sea level, causing the
boiling point to slightly rise or fall depending on your geographical location. B. [(102–
100)/100]*100 = 2% [(99.2–100.0)/100]*100 = –.8% C. 1.20 g/mL D. 1.35 cm3 E. The
measurement would be incorrect as some of the water could potentially splash out of the beaker. F.
Calculated volume gives a more accurate measurement as finding the volume of the displaced water
can be more difficult than making errors while taking dimensions. G. The density of the sample is
about 15 g/cm3. From this, we can draw the conclusion that it is not real gold, as its density is about
19.30 g/cm3. H. In order to prepare 10 mL of 0.25 M HCl solution, you need dilute 2.5 mL of 1M
HCl to 10 mL by adding (10 mL – 2.5

CHM130 Lab 6
CHM130
Lab 6
Exploring Density
Name: Paige Miller
A. Data Tables
Place your completed Data Tables here:
Part IIIa (3 points)
Volume of water in graduated cylinder (mL) 10.00 mL
Mass of rubber stopper (g) 11.37 g
Volume of water and rubber stopper (mL) 16.50 mL
Part IIIb (6 points)
Volume of water in graduated cylinder (mL) 20.00 mL
Mass of iron nail (g) 3.45 g
Volume of water and iron nail (mL) 20.50 mL
Part IV (20 points)
Type of Aluminum Foil
Mass (g)
Length (cm)
Width (cm)
Volume (cm3)
Thickness (cm)
Regular
0.67 g 15.03 cm 10.02 cm 1.81 cm3 Heavy Duty 0.97 g 15.02 cm 10.01 cm 2.62 cm3 B. Follow Up
Questions
Show all work for questions involving calculations.
Part I
1. Use the concepts/vocabulary of ... Show more content on Helpwriting.net ...
Now that you know which is more dense, what must be true of the particles that make it more
dense?
– Prior to completing this lab I believed the nail would be more dense than the stopper. After
performing the experiment, the iron nail contains more particles by volume than the rubber stopper
does.
Part IV
10. The accepted value for the density of aluminum is 2.699 g/cm3. Using this density and the
masses of your two samples of aluminum foil, calculate the volume of aluminum in each sample:

the regular vs. the heavy duty. (Note: You will have to use a bit of algebra and the formula D = m/v
for this calculation!) You must show all your work. (15 pts)
– Regular Aluminum Foil:
2.699 g/cm3=0.67g/volume
2.699 g/cm3 x 0.67g= volume
Volume= 1.81 cm3
–Heavy Duty Aluminum Foil:
2.699 g/cm3=0.97g/volume
2.699g/cm3 x 0.97g=volume
Volume= 2.62 cm3
11. Using the volume(s) you just calculated for regular vs. heavy duty samples as well as your
dimensional measurements (length and width in cm) from Part IV of this experiment, calculate the
height, or thickness, of each sample of aluminum using the formula V = l x w x h. In the formula, V
stands for volume, l for length, w for width, and h for height. Once again, you will have to use your
algebraic skills to manipulate the formula, to solve for height. You must show all your work. (15 pts)
– Regular Aluminum Foil:
1.81 cm3= 15.03 cm x

Physics Assignment for Class Xi
Physics assignment :: XI
1. A vertical U tube of uniform cross section contains mercury in both of its arms. A glycerine (d =
1.3 g/cm3) column of length 10cm is introduced into one of the arms. Oil of density 0.8g/cm3 is
poured in the other arm until the upper surfaces of the oil and glycerine are in the same horizontal
level. Find the length of oil column. Density of mercury is 13.6 g/cm3. (9.6cm)
2. Two communicating cylindrical tubes contain mercury. The diameter of one vessel is four times
larger than the diameter of the other. A column of water of height 70cm is poured into the narrow
vessel. How much will the mercury level rise in the other vessel and how much will it sink in the
narrow one? How much will the ... Show more content on Helpwriting.net ...
a) 2.2 g b) 4.4 g c) 1.1 g d) 3.6 g
15. A large wooden piece in the form of a cylinder floats on water with two–thirds of its length
immersed. When a man stands on its upper surface, a further one–sixth of its length is immersed.
The ratio of the masses of the man and the wooden piece is a) 1 : 2 b) 1 : 3 c) 1 : 4 d) 1 : 5
16. An ice cube is floating in water above which a layer of a lighter oil is poured. As the ice melts
completely, the level of interface and upper of layer of oil surface a) rise and fall b) fail and rise c)
not change and not change d) not change and fall
17. The spring balance A reads 2 kg with a block suspended from it. A balance B reads 5 kg when a
beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the
hanging mass is inside the liquid in the beaker as shown in the figure. In this situation, a) the
balance A will read more than 2 kg b) the balance B will read more than 5 kg c) the balance A will
read less than 2 kg and B will read more than 5 kg d) the balances A and B will read 2 kg and 5 kg
respectively
18. Equal volumes of liquid are poured in the three vessels A, B and C (h1 < h2 < h3). All the
vessels have same base area. Select the correct alternatives.
[pic]
a) The force on the base will be maximum in vessel A. b) The force on the base will be maximum in
vessel C. c) Net force exerted by the liquid in all the

Small Organism Worksheet Essay
Name:
Worksheet to investigate the influence of organismal size and shape on the ratio of surface area to
volume.
Complete the following exercises.
Part 1 Let's investigate how size influences surface are to volume ratios. Let's assume we have a
cube with a volume of 1cm3. Each side of the block equals 1cm. Picture a die for each cube.
1. How many sides does a cube have?
2. Let's use 1 block to represent a small cubed organism and 8 blocks to construct a large cubed
organism. Complete the following table. Total Volume Total surface area Surface area:volume ratio
(= # of blocks) (# of surfaces exposed)
_______________________________________________________________________________
Small Organism
_______________________________________________________________________________
... Show more content on Helpwriting.net ...
When comparing two organisms of similar shape, which size organism (large or small) will retain
more heat in a cold environment? Hint: Where is heat typically lost?
Part 2 let's investigate how organismal shape can influence the ratio of surface area to volume.
Use 8 blocks to construct a large rectangular organism and compare it to the cubed organism from
the first exercise. Total Volume Total surface area Surface area:volume ratio (= # of blocks) (# of
surfaces exposed)
_______________________________________________________________________________
Large cubed
_______________________________________________________________________________
Large rectangular
_______________________________________________________________________________
Use this date to answer the following questions
1. I find that there are only _________ units of surface area in the cubed organism and _________
Units of surface are in the rectangular organism. Therefore, rectangular organisms have a (larger or
smaller) surface area : volume ratio than cubed organisms.

2. Which shape organism (rectangular or cubed) would be able to diffuse gases more rapidly through
the surface of its body.
3. Which shape organism (rectangular or cubed) would retain more heat in a cold

Diffusion Investigation
Data Analysis:
With this experiment, the affect of surface area to volume ratio on the rate of diffusion is
investigated. Four agars containing phenolphthalein, an indicator that turns colorless in acid, each
ranging in size were used. In this experiment, the agar cubes represent cells and the time it took for
cubes to turn colorless shows the rate of diffusion. Next, the cubes were placed in a beaker and just
enough hydrochloric acid was added to cover them. Then, the time it took them to turn colorless was
measured. It was observed that when volume increases, so does the surface area, but more so. For
example, the volume of the smallest cube was 0.125cm3 while it's surface are was 1.5 cm2. The
difference of volumes between the smallest and ... Show more content on Helpwriting.net ...
Most importantly, it's very hard to give exact measurements for the agar cubes. Therefore, each
group could have had varying sizes of cubes or the size could have differed from the one we said it
was. For example, the cube with the dimension of 1cm could have been more or less than that. To
improve the accuracy, we can use an electronic balance to make sure that the cubes with the same
dimensions have the same mass to ensure that their sizes are equal. Moreover, we didn't measure the
amount of hydrochloric acid we put into the beakers within the groups as the only instruction we got
was to pour it in until the cubes are covered. Having varied amounts of hydrochloric acid became a
source of error in the experiment. Next time, we should put the same amount of hydrochloric acid
into the beakers. We can do this by using a measuring cylinder. Furthermore, not taking into account
the possibility of an outlier became a source of error in the experiment. Not taking out the outlier
from the data or repeating it for that specific one will wrongly influence the average, which in turn
will influence the standard deviation. This could lead us to think that the experiment was more or
less reliable than it actually was. For example, for the cubes with 1cm dimension, one of the values
was 209 seconds while the others were 596,790, 720,750 and 841 seconds. This suggests that an
error has been made in that

Experiment Two: Laboratory Techniques And Measurements
Experiment 2: Laboratory Techniques and Measurements
Course Number: CHE 111
Abstract: This experiment introduced the student to lab techniques and measurements. It started with
measuring length. An example of this would be the length of a nickel, which is 2cm. The next part
of the experiment was measuring temperature. I found that water boils around 95ºC at 6600ft. Ice
also has a significant effect on the temperature of water from the tap. Ice dropped the temperature
about 15ºC. Volumetric measurements were the basis of the 3rd part of the experiment. It was
displayed during this experiment that a pipet holds about 4mL and that there are approximately 27
drops/mL from a short stem pipet. Part 4 introduced the student to measuring ... Show more content
It used mass, temperature, length, volume, density, and making a dilute solution. I learned the
importance as well as the difficulty of making proper measurements in a lab setting. If one
measurement is off, it will throw the entire equation off. This will give either incorrect or inaccurate
results.
Questions:
A. Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what
could be the reason? The experiment was conducted at 6600ft. Water tends to boil at a lower
temperature at higher altitudes.
B. While heating two different samples of water at sea level, one boils at 102°C and one boils at
99.2°C. Calculate the percent error for each sample from the theoretical 100.0°C.
102°C–100°C/100°C*100%=0.02% 99.2°C –100°C /100°C *100%= –0.01%
Percent error = 0.02% and –0.01%
C. An unknown, rectangular substance measures 3.6 cm high, 4.21 cm long, and 1.17 cm wide.
If the mass is 21.3g, what is this substance's density? 21.3g/17.73cm³=1.20g/cm³
D. A sample of gold (Au) has a mass of 26.15 g. Given that the theoretical density is 19.30 g/mL,
what is the volume of the gold sample? 26.15g/19.30g/mL=1.35mL
E. Which method for determining density is more accurate, the water displacement method or the

Archimedes' principle method? Why? The water displacement method is more accurate. The string
used to suspend the object in the Archimedes' Method could

Essay on Lab Technique and Measurments
Lab Jessica Cimaroli Lab 1 Purpose To learn about the International System of Units (SI) system
and how it relates to measurements in mass, length, temperature, volume, and time. To learn about
the common techniques and laboratory equipment used to make SI measurents. Procedure Length
Measurements 1. Gather the metric ruler, CD or DVD, key, spoon, and fork. 2. Look at the
calibration marks on your ruler to determine the degree of uncertainty and number of significant
figures that can be made when measuring with a ruler. 3. Measure the ... Show more content on
Helpwriting.net ...
5. Calculate the mass of the water by subtracting "Mass A" from "Mass B." Record the mass of the
water in Data Table 4. 6. Pour the water down the drain and fully dry the graduated cylinder. 7.
Repeat steps 2 through 6 for the isopropyl alcohol. 8. Calculate the densities of both the water and
the isopropyl alcohol and record in Data Table 4. 9. The accepted value for the density of water is 1
g/mL and the accepted density for isopropyl alcohol is 0.786 g/mL. Determine the percent error
between your calculated densities and the accepted values for both water and isopropyl alcohol.
Record the percent error in Data Table 4. Volume and Density Measurements (Solid) 10. Gather the
metal bolt, string, magnet, graduated cylinder, beaker, metric ruler, and scale. 11. Tare the scale by
pressing the Φ/T button so that the scale reads 0.0 g. 12. Place the magnet on the scale to measure
the mass of the object. Record the mass in Data Table 5. 13. Use the ruler to measure the length,
width, and height of the magnet in centimeters. Record the measurements in Data Table 5. 14.
Calculate the volume of the magnet by multiplying the length × width × height, record in Data Table
5. 15. Calculate the density of the magnet by dividing the mass

Assignment 7: Analysis Of Results
Question 7: Analysis of Results : After my experiment I take the mass and volume from the small
pieces of clay from all three trials and I find the average which is 14.7g (mass) and 8.7cm3
(volume). Then I divide using thee formula d=m/v and for density I got 1.7g/cm3, rounded to the
nearest tenth. Next I did the same for medium of all three trials in which I got 22.5g for the average
of the mass and 13.3cm3 for the volume. I divide using the same formula and also got 1.7g/cm3,
rounded to the nearest tenth. Lastly I find the average of mass and volume for large of all 3 trials and
got 40.1g (mass) and 24.0cm3 (volume). I divide 40.1g and 13.3cm3 and got 1.7g/cm3 for the
density, rounded to the nearest tenth. I compare the mass and the difference was 0 (when all are
rounded to the nearest tenth) which shows size doesn't really affect density. ... Show more content
I concluded from my results that my hypothesis is correct. In my analysis I said that, I divided the
average of the mass and volume and I got 1.7g/cm3 for each clay's density. Also in the article:
Physical Properties: Density by Martha Day and Anthony Carpi it states "Because it is a ratio, the
density of a material remains the same no matter how much of that material is presented." Using
these results and information I found out that volume/size doesn't affect

Cellular Organisms: Large Surface Area To Volume Ratio
Multi cellular organisms have a much smaller surface area to volume ratio. Many of their cells are
not in contact with their surroundings so they con not only rely on diffusion to supply all their
organs with oxygen and nutrients, as the distance from their surface to all cells is too fare. We are
multi cellular and have special surface for gaseous exchange and for obtaining nutrients. However
Single celled organisms are small, which means that their surface area is large compared with their
volume; they have a large surface area to volume ratio. Therefore they can obtain substances by
diffusion through their relatively large plasma membrane. These substances have to diffuse only
short distance so can diffuse at a faster rate and meet the organism's ... Show more content on
Helpwriting.net ...
If a cell has a very large surface area and a small volume, this would suggest that there is more cell
membrane (surface area) through which diffusion, osmosis and active transport may take place, and
the substances will reach the internal volume of the cell quickly. Moving substances quickly into or
out of a cell is critical for life processes. If food takes too long to get across the membrane and reach
the internal volume, then the cell will starve. Similarly, if toxins cannot be removed out of the cell
quickly, and accumulates in its internal volume, the cell will die. Large multi cellular organisms like
us have very small surface area to volume ratio. There is a lot of volume that forms the size of our
bodies, but the surface area of our skin that cannot provide for the rapid transport of materials into
and out of the deepest recesses in our bodies. Cells that are buried under layers of other cells cannot
rely on simple diffusion, osmosis or active transport for the exchange of substances simply because
this will take too long. Therefore, to increase surface area to volume ratio, multicellular organisms
develop complex systems of tubes and a channel that delivers solutions from their external
environment to the cells buried deep inside them. This system of tubes and channels becomes what
is known as the transport system in these

Physical Properties Of Matter, Density, Mass, And Volume
One of the physical properties of matter is density. Density is defined as a measure of how much
mass is contained in a given unit volume. A physical property can be measured without changing the
chemical identity of the substance. Since pure substances have unique density values, measuring the
density of a substance can help identify that substance. Density is determined by dividing the mass
of a substance by its volume. Density, volume, and mass have a clear relationship that can clearly be
shown by the mathematical formula:
Density= Mass/Volume
In our experiment we used the determined mass and volume of the sample to calculate the density of
an irregular solid and of an unknown liquid. We then compared our density of our liquid with
densities of known substances to determine the identity of the unknown substance. The purpose of
the experiment was to determine the density of a solid and an unknown liquid. Furthermore, we also
evaluated the relationships of density, mass, and volume.
Keywords: Density, Mass, Volume
Introduction
"Which is heavier, a pound of feathers or a pound of lead?" I'm sure we have all heard this pretty
naive riddle. However, some people would think that the answer would be the pound of lead.
However, after this experiment we should all know the answer.
All matter has volume and mass. Density expresses the relationship of these two primary properties
of matter. Mass is the measure of how much matter an object contains. The more matter an object

Investigating Correlation Between A Solution Salt...
The objective of this experiment is to determine the correlation between a solution's salt
concentration and its density. Also, it was to determine what equipment would be best to use to
measure volume precisely and accurately. Densities of a series of known salt solutions were trialed
three times and then averaged. By following the procedures, the unknown salt solution was
determined and graphed to calculate the percent salt to the sample.
Introduction
In chemistry there are often times an unknown solution's density must be determined by using the
known solutions. There are a variety of modern ways that chemist use to determine an unknown, but
the simplest method to find the identity is by using a common equation: p=M/V P is the density, M
is the mass, and V is the volume. Once the densities of the compounds are known, the density of the
unknown solution can be compared to the four known solutions. When being compared the
solutions can be graphed by plotting density against the percentages of NaCl. Once graphed there
will be an evident straight–line graph.
Experimental
General Methods and Materials.
Materials:
1. NaCl solutions of the following strengths: 6% 12% 18%, 24%
2. Unknown brine solution
3. Volumetric flasks, small beakers, droppers, 10 mL and 50 mL graduated cylinders
4. Scale that measures mass in grams
Procedure. For this experiment, the first task was to determine which equipment would be used to
measure volume accurately. The choices were a 10.0 mL

Solid Mensuration
CPR
(MATH13– B10)
Members: C06 Wrenbria Ngo C07 Julie – Ann Parañal C08 Dani Patalinghog C09 Marino Penuliar
C10 Michael Sadsad
CPR
(MATH13– B10)
Members: C06 Wrenbria Ngo C07 Julie – Ann Parañal C08 Dani Patalinghog C09 Marino Penuliar
C10 Michael Sadsad
Prof. Charity Hope Gayatin
Prof. Charity Hope Gayatin
Homework 1.1
#15. Find the sides of each of the two polygons if the total number of sides of the polygons is 13,
and the sum of the number of diagonals of the polygons is 25.
Assume: ... Show more content on Helpwriting.net ...
Find the lengths of the three sides if the area of the triangle is 576cm2 .
Soltion : c/17=9/10=b/9
A= ss–as–b(s–c) s= a+b+c2 s= 10a+9a+17a20 b= 9(40)10 s= 95a b=36
576= 18a250; c= 17(40)10
9a2= 14400 c= 68 a2= 1600 a=40 Answer: 40cm, 36cm, 68cm
#15. Given triangle ABC whose sides are AB=15in., AC=25 in., and BC= 30in. From a point D on
side AB, a line DE is drawn to a point E on side AC such that angle ADE is equal to angle ABC. If
the perimeter of triangle ADE is 28 in., find the lengths of the line segments BD and CE. Given: ? A
D 30in 15in
B E C
Required: BD =? ; CE =?
Solution: For BD P ADEP ABC = ADAB P ADE=28in Answer: The length of segments BD and CE
is 9in and 10in AD = 15in( 28in)70in P ABC=70in AD = 6in P ADEP ABC= AEAC BD = 9in AE =
25in (28in)70in AE = 10 in

#17. What is the sum of the areas of the two triangles formed in number 16? Given: 3

Effect of Concentration on Enzyme Activity Lab Report
Lab Report
(Effect of concentration on enzyme activity)
Biology
Noor Alawadhi
11– KC
Introduction:
An Enzyme is a protein, which is capable of starting a chemical reaction, which involves the
formation or breakage of chemical bonds. A substrate is the surface or material on or from which an
organism lives, grows, or obtains its nourishment. In this case it is hydrogen peroxide. This lab
report will be explaining the experiment held to understand the effects of the changes in the amount
of substrate on the enzyme's reaction.
Research question:
What does the changes in the amount of substrate on an enzyme's reaction effect on?
Hypothesis:
I forecast that the more concentrated the hydrogen peroxide is the higher the volume of ... Show
Make sure you rinse your glassware carefully between procedures. Repeat this procedure five times.
5. Use this data to calculate the rate of reaction of this enzyme 6. Plan and conduct your own
investigation into one other factor that affects the activity of catalase, you may use this as the basis
for your investigation but if it is to be for moderation of internal assessment you must modify it
substantially.
Results: Different type of hydrogen peroxide | Volume of foam cm³ | 0.5% | 5 cm³ | 1% | 7 cm³ | 2% |
13 cm³ | 4% | 21 cm³ | 6% | 24 cm³ |
The volume of the foam using different concentrations of hydrogen peroxide:
Conclusion:
As I forecasted the results proved correct that, the more concentrated the hydrogen peroxide is the
higher the volume of the foam is and the less concentrated the hydrogen peroxide is the lower the
volume of the foam will become. Other than the table of results the graph is also evidence to my true
hypothesis. Each time the concentration of hydrogen peroxide doubles the level of foam increases

but the more the concentration the foam reaches its constant point therefore slowly increasing. An
example is when the 2% concentration of hydrogen peroxide was added the foam level was 13cm³
but when 4% was added to the next trial the level of foam greatly increased to 21 cm³ and as it
reached its constant point, when 6% it slowed down and increased only 3 cm³ more which became
24 cm³.
Evaluation:

Salt Water Lab Report
We did Lab 2 part 1, first we found the weight of the graduated cylinder and 10 ml of water. Then
we found the weight of the graduated cylinder. After that we found the we found the weight of the
10 ml of water by subtracting the weight of the water and the cylinder by the weight of only the
cylinder. We found the volume of the water by seeing how much the water rises but it did not rise
from 10 ml mark because no object or weight was in the water. The density of the water is mass
divided by volume. Our volume was 1 ml and our, mass was 9.8g, and after dividing those two
numbers we got a density of .98g/ml. But the density of pure water is 1.0, and the density we got for
the density of tap water which is not pure water, is supposed to weigh more than the pure water. In
Lab 2 part 3 the wt of the 10 ml of salt water and G.C together was 37g about 1g difference from the
weight of 10 ml of H2O and the G.C. The weight of the G.C alone is 27g. The weight of the salt
water alone is 10g, the volume of it was 10ml. The density of the salt water was 1g/ml. The
discrepancy in this data about part 1 and part 3 of lab 2.The data of the 2 parts only have a
difference of 1g or lower.
The amount of fluid pushed aside by an object, the buoyancy depends on the space an object takes
up, ... Show more content on Helpwriting.net ...
We also use density for cooking. Steam, if it was dense than we would not be able to take the heat
away from the food or object. We also use density for transportation like boats. The boats depend on
density to stay afloat. Density also helps us with regular activities like swimming, density helps us
stay afloat when we swim. Lincoln had a idea with his proposal to use buoyant force in protecting
boats from running aground. Buoyancy keeps things afloat. A fish has gas in its internal bladder
which makes the fish float instead of

Essay On Density Determination Lab
Hypothesis
The objective of the density determination lab is to determine/compare the average density of pre–
1982 and post–1982 pennies. This can be achieved by finding the mass to volume ratio then
averaging the results. The purpose of this lab is to find out if the difference in composition of the pre
and post pennies will affect their densities. Pennies made before 1982 where majority copper but
pennies minted after 1982 are made with mostly zinc. With this background knowledge, we expect
the outcome to be that the density of the pre–pennies will be greater than the post–pennies. This
experiment will help us to determine if the pre pennies will have a higher density by finding the
volumes and mass's then comparing the results. ... Show more content on Helpwriting.net ...
Calculations
The starting volume of water was 25 ml. To find the volume of the pennies by itself, the volume of
water/pennies together needs to be subtracted from the volume of the water. To find the density the
equation D=M/V must be used. The mass of five post 1982 pennies is 12.37g. The total volume of
five post pennies and the water is 26.5ml. The volume of five post pennies alone is 1.5ml. 26.5ml–
25ml=1.5ml The density for five post pennies is 8.25g/ml. 12.37g/1.5ml=8.25g/ml The mass of ten
post 1982 pennies is 25.59g. Ten post pennies and water have a volume of 28.5ml. Ten pennies
volume alone is 3.5ml 28.5ml–25ml= 3.5 The density of ten post pennies is 7.31g/ml.
25.59g/3.5ml=7.31g/ml The mass of fifteen post pennies are 38.21g. The volume of fifteen pennies
and the water is 30.5ml and the pennies volume alone is 5.5ml. 30.5ml–25ml=5.5 The density of
fifteen post pennies is 6.95g/ml. 38.21g/5.5ml=6.95g/ml The mass of twenty post pennies is 50.78g.
The volume of twenty pennies and water together is 32.5ml and the post pennies alone are 7.0ml.
32.5ml–25ml=7.0. The density of twenty post pennies is 7.25g/ml. 50.78g/7.0ml=7.25g/ml. The
average density of pre 1982 pennies is 7.44 g/ml. 8.25g/ml + 7.31 g/ml + 6.95g/ml + 7.25g/ml=
29.76g/ml, 29.76g/ml / 4= 7.44 g/ml
The volume of water started with is 25 ml. The mass of five pre 1982 pennies is 14.2g. The total
volume of five

Fermentation Lab Report
Fermentation is the anaerobic process by which glucose, or other sugars are catabolized by
microorganisms without an electron transport chain, like yeast (Campbell, 2004). In experiment one
the yeast and glucose acted as reactants, with the yeast breaking down the glucose, producing the
products CO2 and ethanol. The overall reaction for the alcoholic fermentation that took place can be
represented as 2 pyruvate + 2NADH–> 2NAD+ +2CO2 + 2 ethanol (Campbell, 2004). This
biological process allows cells to operate under conditions where oxygen is not present. Experiment
1 explored the question of how the amount of sugar impacts the rate of fermentation, while
Experiment 2 tested the effect the type of sugar available to the yeast has on the rate of fermentation.
If more sugar is available to the yeast, the faster the reaction should occur. If sucrose is used instead
of glucose, the reaction will produce less CO2 since the yeast cannot breakdown sucrose as
efficiently as glucose.
Methods:
The procedure for experiment one followed that given in: Martineau, Dean, Gilliland, Soderstrom.
BIO 19L Experiment 1, General Biology I Lab for Science Majors, Autumn Quarter, 2017 – 2018.
DePaul University.
Experiment one tested the effect the amount of sugar has on the fermentation rate. Four labeled 50
milliliter (mL) test tubes were used to transport each reaction to the designated fermentation tube.
Each tube had a specific ratio of reactants. Test tube 1 acted as the negative control and

Water Density Lab
Density is the measurement of an object that has a relative mass and is divided by a specific volume.
Mass is the amount of matter an object possesses. Volume is the amount of space of an object
occupies (solid, liquid or gas.) In chemistry the term density can be described as a physical and
intensive property of matter. The term intensive property means, it is independent on the size and
amount of a substance. Many chemist compare different types of densities of elements and
substance with water. Density can be either expressed as (g/mL) for liquids, (g/cm3) for solids, and
grams per liter (g/L) for gases. (Ch. 7 Notes, Dr. Binz, Lindenwood University, 2015.) The density
of water is about 1.00 g/mL. If the density of an object is less ... Show more content on
Helpwriting.net ...
It can be calculated by determining the average mass and average volume of the known and
unknown zodiaq quartz. A new method that was learned from doing the experiment was being able
to measure mass and volume which later helps scientist identify a either a new complete substance
or even just an unordinary substance.
The percent error in this experiment was roughly under six percent, which was appropriate in this
experiment. One reason for high percent error was when the mass was being measured, the scale
was need to be zeroed. Once it was zeroed, the scale must be completely motionless. So the mass of
the weight could have been measure when the scale was not still, which meant it was not accurately
zeroed. The final reason for the high percent error was when the object was placed on the weight,
the table was begin to shake due to others surrounding the table. This could have created issues,
when zeroing the scale at the beginning, and when trying to find the exact mass of the object. The
table was not perfectly still, so this could have been an issue when determining the mass for each

Laboratory Techniques and Measurements Essay
Lab #3 January 30, 2012 1. Title: Laboratory Techniques and Measurements 2. Purpose: The
purpose of this lab is to learn laboratory techniques and to how to measure precisely. During this lab
I will learn how to measure length, temperatures, volume, density, and mass using laboratory
equipment. I will be using laboratory equipment to prepare dilutions and calculate them while using
an algebraic formula. 3. Procedure: In order to perform this lab I needed the following materials: a
metric ruler, table salt, small number of ice cubes, piece of string, flame source, rubbing alcohol, tap
water, distilled water, paper, colored liquid drink, 100mL glass beaker, burner–fuel, burner–stand, 25
mL cylinder, 25mL volumetric ... Show more content on Helpwriting.net ...
Observe and note the colour and intensity of the drink (solution). The solution's coloring prior to
adding water was a bright red color. b. Observe and note the color and intensity. After adding the
distilled water to the solution the color starts to slowly dim to a lighter, and lighter red. Towards the
end of adding droplets the solution surprised me because the coloring actually separated. As the

How Does the Surface Area to Volume Ratio Affect Heat Loss...
Biology Lab Report
BY:
Michael Ryan Pranata 11C |
Background: As heat is a form of thermal energy, they tend to have the behavior of reaching a
thermal equilibrium. This means that when two bodies of different temperatures come in contact
with each other, the hotter ones will transfer heat particles to the body with a colder temperature,
with an aim to reach this "thermal equilibrium", whatever the temperature may be. The larger the
surface area, means there can be more "paths" from the sides of the body that are capable of
releasing this heat particles, and reaching thermal equilibrium faster. This is what happens when a
hotter body is subjected to a colder one.
Research Question:
How does the surface area to volume ratio ... Show more content on Helpwriting.net ...
We observe the temperature of the four different flasks starting from the equal initial temperatures,
every ten seconds for one samples all the way to three minutes. From the data observed, the average
rate of heat loss can be derived. This is done by taking the initial and final temperature and dividing
them by the number of seconds (18x as the total number of seconds is 180 and interval of 10 secs).
Controlled: Initial temperature, type of insulation, type of flasks, room temperature, time.
1. We've decided 77⁰C as our initial temperature for every trial. We've kept the initial temperature
equal for each trial as this would give us fair and reliable results. If the initial temperatures were
kept different, therefore the data provided will be somewhat unreliable. This is also because as the
temperatures are higher, the rate of heat loss would be faster, as compared to when the initial
temperatures are lower. This would again support the unreliability of the data we would observe.
2. The type of insulation that we've chosen is the cotton, approximately with the same thickness for
each flask. They will be changed for every trial, because for each trial, as the heat escapes from the
sides of the flask in the form of water vapor, it comes in contact with the cotton, thus making it
moist, and changes the capacity of the cotton to insulate. We have decided to

Lab Report Density Lab

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