There are two ways to solve this problem. I will .pdf
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There are two ways to solve this problem. I will show both ways. First, you can find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14. So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take - log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also, notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this helps and please rate. :) Solution There are two ways to solve this problem. I will show both ways. First, you can find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14. So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take - log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also, notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this helps and please rate. :)..
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![There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)
Solution
There are two ways to solve this problem. I will show both ways. First, you can
find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the
concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract
both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve
this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14.
So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take -
log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also,
notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this
helps and please rate. :)](https://image.slidesharecdn.com/therearetwowaystosolvethisproblem-230629200117-5189c7c4/85/There-are-two-ways-to-solve-this-problem-I-will-pdf-1-320.jpg)
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Complete the following table by calculating the missing entries and indicating whether the solution is, acidic or basic. (Make sure you show your work!) OH1 Acidic or basic? [H\'1 pH pOH 6.8 x 10-3 M 2.1 x 109 M 6.73 4.25 HTML Editor OH-] Acidic or Basic? pH pOH 6.8 x 103 M 2.167 2.1 x 109 M 6.73 4.25 table tbody tr tod 19 words
Solution
[H+] = 6.8x10^-3M
[H+][OH-] = Kw where Kw = ionic product of water = 1.0x10^-14
[OH-] = Kw/[H+] = 1.0x10^-14/6.8x10^-3 = 1.47x10^-12
[OH] = 1.47x10^-12M
[H+] = 6.8x10^-3
-log[H+] = -log( 6.8x10^-3)
PH= 2.167
PH+POH= 14
POH= 14-PH
POH= 14-2.167 =11.833
[H+] = 6.8x10^-3M, [OH-]= 1.47x10^-12M, PH= 2.167 , POH= 11.833
PH<7, so the nature of solution is acidic.
b) [OH-] = 2.1x10^-9M
[H+] = 1.0x10^-14/2.1x10^-9 = 4.76x10^-6
[H+] = 4.76x10^-6M
-log[H+] = -log( 4.76x10^-6)
PH= 5.322
POH= 14-5.322
POH= 8.678
[H+] = 4.76x10^-6M , [OH-] = 2.1x10^-9M , PH= 5.322, POH= 8.678
PH<7, so the nature of the solution is acidic
c)
PH= 6.73
-log[H+] = 6.73
[H+] = 10^-6.73
[H+] = 1.86x10^-7M
[OH-] = 1.0x10^-14/1.86x10^-7
[OH-] = 5.376x10^-8M
POH= 14-6.73
POH= 7.27
[H+] = 1.86x10^-7M. [OH-] = 5.376x10^-8, PH= 6.73, POH= 7.27
PH<7, so the nature of the solution is acidic nature,
d)
POH=4.25
PH= 14-4.25=9.75
PH= 9.75
-log[H+] = 9.75
[H+] = 10^-9.75
[H+] = 1.778x10^-10M
[OH-] = 1.0x10^-14/1.778x10^-10
[OH-] = 5.62x10^-5
[H+] 1.778x10^-10M, [OH-] = 5.62x10^-5M , PH= 9.75, POH= 4.25
PH>7, so the nature of the solution is basic .
.
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Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
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electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
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operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
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temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th..
This is a titration of a weak base w a strong ac.pdf
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation.
Solution
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation...
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of elements in a molecule, not the actual number of atoms. The empirical formula for the molecule described is A1Br3, indicating it contains aluminum, bromine, and the subscripts show their relative amounts in the molecule. The empirical formula provides the simplest whole number ratio of the elements in a compound.
More Related Content
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1) Identifying conjugate acid-base pairs from chemical equations.
2) Explaining the pH scale and how it relates to hydrogen ion concentration.
3) Demonstrating calculations for pH, pOH, and hydrogen/hydroxide ion concentrations using mathematical relationships like pH = -log[H+].
P H Scale
The document discusses the pH scale and how it is used to measure acidity and basicity in solutions. It defines pH as the negative log of the hydrogen ion concentration. Solutions with pH < 7 are acids, pH = 7 are neutral, and pH > 7 are bases. It also discusses how water can act as both an acid and base and how the pH, pOH, [H+], and [OH-] of any solution are related through water's ionization constant, Kw.
Ph and POH
This document provides an overview of acid/base chemistry and pH. It defines pH as a measure of hydrogen ion concentration, describes the pH scale from 0-14, and explains how to calculate pH, pOH, and hydrogen or hydroxide ion concentration from other values. Sample problems demonstrate how to determine pH from concentration and vice versa, as well as the relationship between pH and pOH. Key points are that pH is a log scale measurement of acidity, and that the sum of pH and pOH equals 14 for any aqueous solution.
How do we measure acids and bases
This document discusses pH and how it is used to measure the acidity or basicity of solutions. It explains that pH is determined by measuring the concentration of hydrogen ions [H+] or hydronium ions [H3O+] in a solution. The pH scale ranges from 0 to 14, with acids having a pH below 7, bases above 7, and neutrals at 7. It also provides examples of calculating the pH of acid and base solutions given their concentrations.
Acid base titrations
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2) Key aspects include calculating the equivalence point and determining the pH in four regions: before, at, and after the equivalence point as well as in the buffer region between initial addition and the equivalence point.
3) Factors that affect the titration curve include the strength of the acid and its concentration - weaker acids and more dilute solutions result in shallower curves without a clear endpoint.
asam dan basa
This document provides an overview of acid-base concepts including definitions, examples, and calculations. It defines acids as producing H+ ions and bases as producing OH- ions. Strong acids and bases fully dissociate in water while weak acids and bases partially dissociate. Acids are proton donors and bases are proton acceptors. The document also discusses using pH and pOH scales to calculate hydrogen and hydroxide ion concentrations in solutions and determine whether they are acidic or basic.
IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water
This document provides a tutorial on acid-base dissociation constants (Ka and Kb), pKa, pKb, and pH. It defines strong and weak acids and bases, and explains how strong acids fully dissociate in water while weak acids only partially dissociate. Formulas used to calculate pH and concentrations of H+ and OH- are presented. A table shows how the ionic product constant of water (Kw) varies with temperature, and how this affects pH. The document concludes by explaining how temperature impacts acid and base dissociation equilibria.
Ch14 z5e acid base
The document discusses different definitions of acids and bases, including:
1) Arrhenius definitions - acids produce H+ ions in water, bases produce OH- ions. Limited to aqueous solutions.
2) Bronsted-Lowry definitions - acids are H+ donors, bases are H+ acceptors. Acids and bases always come in pairs when reacting.
3) pH scale is used to express acidity because [H+] is usually very small. pH decreases as [H+] increases exponentially. Common substances are classified as acidic, basic, or neutral based on their pH.
Chem unit9
The document discusses various properties of acids and bases, including that acids are sour and donate protons while bases are bitter and accept protons. It also discusses conjugate acid-base pairs, water's ability to act as an acid or base through self-ionization, and the pH scale. Key concepts covered include Brønsted-Lowry definitions of acids and bases, conjugate acid-base pairs, calculating pH and pOH, and neutralization reactions between acids and bases.
Complete the following table by calculating the missing entries and in.docx
Complete the following table by calculating the missing entries and indicating whether the solution is, acidic or basic. (Make sure you show your work!) OH1 Acidic or basic? [H\'1 pH pOH 6.8 x 10-3 M 2.1 x 109 M 6.73 4.25 HTML Editor OH-] Acidic or Basic? pH pOH 6.8 x 103 M 2.167 2.1 x 109 M 6.73 4.25 table tbody tr tod 19 words
Solution
[H+] = 6.8x10^-3M
[H+][OH-] = Kw where Kw = ionic product of water = 1.0x10^-14
[OH-] = Kw/[H+] = 1.0x10^-14/6.8x10^-3 = 1.47x10^-12
[OH] = 1.47x10^-12M
[H+] = 6.8x10^-3
-log[H+] = -log( 6.8x10^-3)
PH= 2.167
PH+POH= 14
POH= 14-PH
POH= 14-2.167 =11.833
[H+] = 6.8x10^-3M, [OH-]= 1.47x10^-12M, PH= 2.167 , POH= 11.833
PH<7, so the nature of solution is acidic.
b) [OH-] = 2.1x10^-9M
[H+] = 1.0x10^-14/2.1x10^-9 = 4.76x10^-6
[H+] = 4.76x10^-6M
-log[H+] = -log( 4.76x10^-6)
PH= 5.322
POH= 14-5.322
POH= 8.678
[H+] = 4.76x10^-6M , [OH-] = 2.1x10^-9M , PH= 5.322, POH= 8.678
PH<7, so the nature of the solution is acidic
c)
PH= 6.73
-log[H+] = 6.73
[H+] = 10^-6.73
[H+] = 1.86x10^-7M
[OH-] = 1.0x10^-14/1.86x10^-7
[OH-] = 5.376x10^-8M
POH= 14-6.73
POH= 7.27
[H+] = 1.86x10^-7M. [OH-] = 5.376x10^-8, PH= 6.73, POH= 7.27
PH<7, so the nature of the solution is acidic nature,
d)
POH=4.25
PH= 14-4.25=9.75
PH= 9.75
-log[H+] = 9.75
[H+] = 10^-9.75
[H+] = 1.778x10^-10M
[OH-] = 1.0x10^-14/1.778x10^-10
[OH-] = 5.62x10^-5
[H+] 1.778x10^-10M, [OH-] = 5.62x10^-5M , PH= 9.75, POH= 4.25
PH>7, so the nature of the solution is basic .
.
Ionic equilibria & p h 1
The document discusses ionic equilibria and pH. It defines pH as the negative logarithm of the hydronium ion concentration in a solution. A pH below 7 indicates an acidic solution, while a pH above 7 indicates a basic or alkaline solution. The document also discusses how dilution affects pH by decreasing concentration and increasing pH. It presents the pH scale from 0-14 and discusses how to calculate pH and pOH values from concentrations. Examples are provided to demonstrate calculating pH for solutions of calcium hydroxide, acetic acid, and the ratio of carbon dioxide to bicarbonate ion concentrations in blood.
IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water
1. Strong acids like HCl dissociate completely in water, generating a high concentration of hydrogen ions (H+), while weak acids like acetic acid (CH3COOH) only partially dissociate, resulting in a lower H+ concentration.
2. The pH scale is a logarithmic measure of hydrogen ion concentration, with lower pH values indicating higher acidity. A change of one pH unit represents a ten-fold change in H+ concentration.
3. The ionic product constant of water (Kw) describes the equilibrium between water and its ions (H+ and OH-). Kw is temperature dependent and increases with rising temperature, resulting in higher concentrations of H+ and OH- ions
CM4106 Review of Lesson 3 (Part 1)
This document provides an overview of additional aspects of acid-base equilibria, including:
1. Important relations for pH, pOH, Ka, and Kb calculations.
2. Steps for calculating pH, pKa, [H+], Ka for acids or bases, including distinguishing between strong and weak acids/bases.
3. Steps for calculating pH for mixtures of acids and bases, including considerations for salt solutions, buffer solutions, and calculating pH changes upon adding small amounts of acid or base to a buffer.
Similar to There are two ways to solve this problem. I will .pdf (13)
IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water
IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water
Complete the following table by calculating the missing entries and in.docx
Complete the following table by calculating the missing entries and in.docx
IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water
IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water
More from ssuseraef9da
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
There are two ways to solve this problem. I will .pdf
The ‘nuclei effect’ Cesium formate brine has a natural high density. The weight
comes from cesium ions rather than added particles, such as barite. In traditional OBMs, the
weighting agent is stopped at the well bore wall and forms a ?lter cake. With cesium formate the
problem of ?lter cakes are eliminated, however one needs to correct for the drilling ?uid that
?ltrates into the near formation. “The traditional logging tool algorithm is only good for elements
with atomic numbers less than roughly 24, which means that the element has one neutron for
every proton. Cesium, with an atomic number of 55, has considerably more neutrons than
protons in its nucleus. Therefore when measuring electron density, which is then used to
calculate bulk density, the algorithm should show an under-estimation of density due to this large
‘nuclei effect,’” explains Pedersen. “The only problem was that bulk density was over-
estimated,” he continues. So what was the explanation? “When going through the data I noticed
an intriguing correlation between the gamma logs and the photoelectric factor. To cut a long
story short, higher-numbered atomic elements, such as cesium, produce a dramatically higher
photoelectric effect. This increases the absorption of gamma rays in the formation so that the
electron density appears higher than it is and physical density is therefore over estimated. This
far outweighs the ‘nuclei effect’ previously mentioned,” says Pedersen. A simple solution Now
StatoilHydro understood why the logging results were wrong when interpreted using
methodology based on drilling ?uids with lower atomic numbers, the challenge was to ?nd a way
of interpreting the numbers accurately. The simplicity of the solution was stunning. After close
examination of the calibration data, a linear relationship was found between the photoelectric
factor (Pef) and the over-estimation of density. Consequently, by consistently adjusting bulk
density by the measured Pef, porosity was correctly calculated. “The beauty of this correction is
that it relies only on the Pef measurements, which are logged simultaneously with bulk density,
and can therefore apply to any cesium formate concentration,” says Pedersen. The perfect model
The logged photoelectric factor isn’t only useful for correcting the density curve and estimating
saturation of the invading ?ltrate, but is also ideal for de?ning permeable sands. Combining the
photoelectric factor and its high vertical resolution with resistivity measurements from both the
drill pass and ream pass produces a very reliable and accurate net reservoir de?nition.
Furthermore, by applying a conductive drilling ?uid based on cesium formate brine in all
production wells high-quality resistivity image logs can be run. These logs provide important
information on structural dip, depositional environment, sedimentary features, facies and
geological correlations, all used by geomodelers to produce better reservoir models. “The results
match cor.
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
This is a titration of a weak base w a strong ac.pdf
This equation makes it able for you to plug-and-play the numbers.
Let\'s classify the variables first.
n is the number of people
p is the profit acquired
If you plug in p=0 (the situation when the profit is 0), you get
0 = 10(120-n)-50n
Now all you have to do is solve for n.
Solution
This equation makes it able for you to plug-and-play the numbers.
Let\'s classify the variables first.
n is the number of people
p is the profit acquired
If you plug in p=0 (the situation when the profit is 0), you get
0 = 10(120-n)-50n
Now all you have to do is solve for n..
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th..
This is a titration of a weak base w a strong ac.pdf
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation.
Solution
This is a titration of a weak base w/ a strong acid. So when you add the acid, it will
dissociate into H+ and Cl- ions completely, but some of A- will consume the H+ and it will form
HA. So this will in affect become a buffered solution. So i believe you can use henderson
hasselbach for this equation...
The empirical formula indicates the relative proportions of the elem.pdf
The empirical formula indicates the relative proportions of elements in a molecule, not the actual number of atoms. The empirical formula for the molecule described is A1Br3, indicating it contains aluminum, bromine, and the subscripts show their relative amounts in the molecule. The empirical formula provides the simplest whole number ratio of the elements in a compound.
1) The main thrust behind the Brexit vote was the fear that UK based.pdf
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment.
Solution
1) Measurement of time depends on the reaction time of the human measuring the time /
operating the stop clock. There will always be error in measuiring the time whereas measuring
the distance is limited by the least count of the measuring device. Hence distance / displacement
measurement yeilds better results compared to experiment invlving measurement of time.
2) Frictional force generated by Air also called drag force which is proportional to the speed and
opposes the motion of the projectile introduces error in the experiment..
The empirical formula indicates the relative proportions of the elem.pdf
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio)
Solution
Samantha Corporation Total Manufacturing OH 490,000 Fixed part of
manufacturing OH 190,000 Variable part of Manufacturing OH 300,000 Total
units produced 100,000 Variable Manufacturing OH /Unit 3 Selling
& Admin OH 520,000 Variable Selling & Admin @$4 /unit 400,000 Fixed
Selling & Admin 120,000 Break even analysis Details Product 1 Product 2
Material cost per unit 6 8 Direct Labor per unit
4 6 Variable Manufacturing OH per unit 3
3 Variable Selling & Admin OH per unit 4 4
Total Variable cost per unit 17 21 Selling Price/unit
24 32 Contribution margin /unit 7
11 Sales volume 60,000 40,000 Sales volume ratio 60%
40% Weighted contribution per unit =7*0.60+11*0.40 Weighted contribution per unit
= 8.60 Total Fixed cost Fixed part of manufacturing OH 190,000
Fixed Selling & Admin 120,000 Total Fixed cost 310,000 Weighted
contribution per unit 8.60 Break Even Units 36,047 Break Even units
Product 1 21,628 (at 60% ratio) Break Even units Product 2 14,419 (at 40%
ratio).
If an oxide has the formula MO, then since O is normally found in th.pdf
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve
Solution
I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance
electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O
+10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that
are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ---->
2HIO3+10NO2 +4H2O Hope this helps, Steve.
1. True2. I am having trouble understanding the phrasing. A precip.pdf
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells.
Solution
1. The answer is: B. Macrophage processes foreign antigen for presentation to T helper cells
which activate B cells to produce antibodies.
3. The answer is: MHC class II markers
MHC class II markers are needed for T helper cell binding to other immune cells..
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
#include algorithm #include vector #include iostream usi.pdf
#include
#include
#include
using namespace std;
template //For Template
void bogosort(std::vector& array) //For bogosort
{
while (! is_sorted(array))
{
std::random_shuffle(array.begin(), array.end());
}
}
template
bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
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Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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- 1. There are two ways to solve this problem. I will show both ways. First, you can find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14. So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take - log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also, notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this helps and please rate. :) Solution There are two ways to solve this problem. I will show both ways. First, you can find the pOH. The formula for pOH is: pOH = -log([OH-]). You take the negative log of the concentration of OH-. pOH = -log(0.15M) = 0.824 Now, pH + pOH = 14. So, you can subtract both sides by pOH to get pH. 14 - pOH = 14 - 0.824 = 13.176 = pH The second way to solve this problem is to find the concentration of H+ ions. We know that [H+] x [OH-] = 1 x 10^-14. So, you can solve for [H+] [H+] = (1 x 10^-14)/(0.15) = 6.67 x 10 ^ -14 Now, you can take - log([H+]) to get pH. pH = -log(6.67x10^-14) = 13.176. Notice you get the same answer. Also, notice when you take -log([OH-]) you get pOH. When you take -log([H+]) you get pH. Hope this helps and please rate. :)