I need help with CH4- What is its intermolecular force- Question 4 of.docx
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I need help with CH4. What is its intermolecular force? Question 4 of 28 Avail Map A Due Point Grad Desc Sapling Learning Identify the intermolecular forces present in each of these substances. Hydrogen bonding dipole-dipole, Policie and dispersion Dipole dipole and dispersion onlyDispersion only You c You c You c until You lo co ??? attem O eText O Help W O Web H O Technic O Previous e?Check Answer 0 Next Exit Solution Dispersion forces present in CH 4 . The only intermolecular forces in methane are London dispersion forces.The major intermolecular forces would be dipole-dipole forces and London dispersion forces. But The electronegativities of C and H are so close that C-H bonds are nonpolar. There are no bond dipoles and no dipole-dipole interactions.Even if the molecule had polar C-H bonds, the symmetry of molecule would cause the bond dipoles to cancel.So the only forces left to consider are London dispersion forces. .
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Hi!For hydrogen fluoride the predominant intermolecular force is .pdf
Hi!
For hydrogen fluoride: the predominant intermolecular force is hydrogen bonding. In hydrogen
bonding, fluorine acts as a \"bridge\" (the middle atom) between two hydrogen molecules. The
only other atoms that can participate in hydrogen bonding are nitrogen, oxygen and sulfur. The
other intermolecular forces are present in hydrogen fluoride as well, but they are not as
significant as hydrogen bonding.
For carbon tetrafluoride: the predominant intermolecular force is dispersion (also called London
dispersion or van der waals forces) due to the fact that this is a non-polar, symmetrical molecule
and the difference in electronegativity between carbon and hydrogen is not great enough for the
bond to be considered truly polar.
For hydrogen chloride: the predominant intermolecular force is dipole-dipole. Chlorine cannot
participate in hydrogen bonding, however, the chlorine atom has a considerable difference in
electronegativity relative to hydrogen so the covalent bond is considered polar. There are also
dispersion forces present, but they are not as significant as the dipole-dipole forces here.
I hope this helps! Please don\'t forget to rate : )
Solution
Hi!
For hydrogen fluoride: the predominant intermolecular force is hydrogen bonding. In hydrogen
bonding, fluorine acts as a \"bridge\" (the middle atom) between two hydrogen molecules. The
only other atoms that can participate in hydrogen bonding are nitrogen, oxygen and sulfur. The
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significant as hydrogen bonding.
For carbon tetrafluoride: the predominant intermolecular force is dispersion (also called London
dispersion or van der waals forces) due to the fact that this is a non-polar, symmetrical molecule
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bond to be considered truly polar.
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I hope this helps! Please don\'t forget to rate : ).
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1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.)
Solution
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
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Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.).
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Okay, let's break this down step-by-step:
* We are given: 1 mole of ice at -25°C
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* Heat of vaporization of water = 40.7 kJ/mol
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* Specific heat of water = 4.18 J/g°C
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2) Heat
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Okay, let's break this down step-by-step:
* We are given: 1 mole of ice at -25°C
* Heat of fusion of ice = 6.01 kJ/mol
* Heat of vaporization of water = 40.7 kJ/mol
* Specific heat of ice = 2.09 J/g°C
* Specific heat of water = 4.18 J/g°C
1) Heat ice from -25°C to 0°C:
Q = m * c * ΔT
Q = (18 g) * (2.09 J/g°C) * (25°C) = 903 J
2) Heat
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14 - (b)
Ether should never be directly kept on a hot plate because it can increase the temperature which is sufficient to start a fire.
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Pinacolone is a ketone, hence it does not react with water( non-polar), HCl, NaOH, aq NaHCO 3 (no acid group) et. However, it gave yellow solution with H 2 SO 4 . H 2 SO 4 being a conc. acid, it chars the compound. Another probable reason is the formation of enol species in acidic medium. Enols are slightly polar, and due to conjugation of OH lone pair with the double bond, the solution becomes coloured.
Me 3 CCOCH 3 + H + = Me 3 CC(OH)=CH 2
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Hi!For hydrogen fluoride the predominant intermolecular force is .pdf
Hi!
For hydrogen fluoride: the predominant intermolecular force is hydrogen bonding. In hydrogen
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I hope this helps! Please don\'t forget to rate : )
Solution
Hi!
For hydrogen fluoride: the predominant intermolecular force is hydrogen bonding. In hydrogen
bonding, fluorine acts as a \"bridge\" (the middle atom) between two hydrogen molecules. The
only other atoms that can participate in hydrogen bonding are nitrogen, oxygen and sulfur. The
other intermolecular forces are present in hydrogen fluoride as well, but they are not as
significant as hydrogen bonding.
For carbon tetrafluoride: the predominant intermolecular force is dispersion (also called London
dispersion or van der waals forces) due to the fact that this is a non-polar, symmetrical molecule
and the difference in electronegativity between carbon and hydrogen is not great enough for the
bond to be considered truly polar.
For hydrogen chloride: the predominant intermolecular force is dipole-dipole. Chlorine cannot
participate in hydrogen bonding, however, the chlorine atom has a considerable difference in
electronegativity relative to hydrogen so the covalent bond is considered polar. There are also
dispersion forces present, but they are not as significant as the dipole-dipole forces here.
I hope this helps! Please don\'t forget to rate : ).
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1) There are several types of intermolecular forces ranging from strongest to weakest - hydrogen bonding, dipole-dipole interactions, dipole-induced dipole interactions, and dispersion forces.
2) Hydrogen bonding is the strongest and involves the electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom like N, O, or F, and another electronegative atom's lone pair of electrons.
3) Dipole-dipole interactions occur between polar molecules that have partially positive and negative ends that electrostatically attract each other, while dipole-induced dipole interactions involve a polar molecule inducing a dipole in a neighboring nonpolar molecule.
topic_4_intermolecular_forces.pdf
The document discusses intermolecular forces and their relative strengths. Hydrogen bonding is the strongest, followed by dipole-dipole forces, then London dispersion forces. Nonpolar molecules experience London dispersion forces while polar molecules experience dipole-dipole forces. Molecules containing hydrogen bonded to oxygen, nitrogen or fluorine experience hydrogen bonding. The type of intermolecular force present determines properties like boiling point, with stronger forces requiring more energy to overcome during boiling.
Lecture 8.4b- Polar Molecules
The document provides information about determining the polarity of molecules based on their structure. It discusses that ethane (C2H6) and ethene (C2H4) have non-polar covalent bonds since the electron sharing is equal, while molecules with unequal electron sharing between atoms with different electronegativity can result in polar bonds. It also explains that a molecule is polar if it has polar bonds and asymmetric structure, but non-polar if it has symmetrical structure or only non-polar bonds. Examples of polar molecules discussed include H2O, NH3, and HCN.
Polar Bonds and Polar Molecules.pptx
This document discusses molecular polarity and how to determine if molecules are polar or nonpolar. It defines polar and nonpolar covalent bonds based on differences in electronegativity between atoms. A molecule is polar if it contains polar bonds in an asymmetrical arrangement, whereas a molecule is nonpolar if the polar bonds are symmetrical or if all bonds are nonpolar. Examples of polar molecules include HCl and H2O, while nonpolar examples include CO2 and CF4.
Chapter 8, section 8.4 (1)
This document discusses bond polarity and molecular polarity. It explains that polar covalent bonds form when electrons are shared unequally between atoms of different electronegativity. The more electronegative atom attracts the electrons more strongly, acquiring a partial negative charge, while the other atom acquires a partial positive charge. Whether a bond is nonpolar, moderately polar, or ionic depends on the electronegativity difference between the atoms. Polar bonds can result in polar molecules if the bond polarities do not cancel out. Intermolecular forces like dipole interactions and dispersion forces are responsible for molecular phase changes and arise from attractions between slightly positively and negatively charged regions on different molecules. Hydrogen bonds are especially strong dipole interactions that
Chemical bond 2.ppt
This document discusses different types of intermolecular forces:
1) Dipole-dipole forces occur between polar molecules due to the attraction between opposite charges.
2) Ion-dipole forces occur between ions and polar molecules.
3) Dispersion forces, the weakest intermolecular forces, arise from temporary dipoles induced in nonpolar molecules.
4) Hydrogen bonding, a stronger intermolecular force, occurs between a hydrogen covalently bonded to F, O or N and another electronegative atom.
L4- Intermolecular Forces of Attraction.pptx
There are three main types of intermolecular forces of attraction: London dispersion forces, dipole-dipole forces, and hydrogen bonds. London dispersion forces exist in all molecules and result from fluctuations in electron distribution. Dipole-dipole forces exist in polar molecules with permanent charge distributions. Hydrogen bonds are strong dipole-dipole interactions between a hydrogen and an electronegative atom like nitrogen, oxygen, or fluorine. These intermolecular forces affect phase changes from solid to liquid to gas as the relative strengths decrease.
IMFA.pptx
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1) A covalent crystal differs from an ionic crystal in that a)on.pdf
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.)
Solution
1) A covalent crystal differs from an ionic crystal in that
a)only one of them is a poor electrical conductor.
b)only one of them has bonding that involves overlap of electronic orbitals.
c)only one of them possesses atomic order.
d)only one of them has high melting points.
e)only one of them is hard.
The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic
crystals involve electrostatic bonding between a positive and negative ion.
2)What are the important intermolecular forces acting in CHF3?
a)London dispersion forces only
b)only dipole-dipole interactions and H-bonding
c)dipole-dipole interactions and London dispersion forces
d)H-bonding, dipole-dipole interactions and London dispersion forces
e)ionic bonding
In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When
Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and
Fluorine are all capable of hydrogen bonding.).
Unit-8-Lesson-15-THE-CHEMICAL-BONDING.pptx
1. The document discusses different types of chemical formulas including molecular, empirical, and structural formulas. It provides examples of each type like H2O for water and C6H12 for hexene.
2. It also discusses ionic and covalent bonding. Ionic bonding involves the complete transfer of electrons from one atom to another, like from sodium to chlorine in NaCl. Covalent bonding involves the sharing of electron pairs between atoms.
3. The document describes electronegativity and how it relates to the polarity of covalent bonds. Polar covalent bonds form between atoms with an electronegativity difference of 0.5-1.6, while ionic bonds form between atoms with a difference above
Gaya Antar Molekul
The document discusses different types of intermolecular forces that influence the physical properties of liquids and solids. It describes three main types of intermolecular forces: 1) dipole-dipole forces including hydrogen bonding, 2) London dispersion forces between all molecules due to induced dipoles, and 3) ion-dipole forces between ions and polar molecules. These intermolecular forces determine how tightly packed molecules are and the strength of attractions between molecules, which in turn influence boiling points, melting points, and whether a substance is a gas, liquid, or solid.
intermolecular_forces.ppt
This document discusses intermolecular forces and the properties they impart to liquids and solids. It describes three main types of intermolecular forces: dipole-dipole forces between polar molecules, London dispersion forces present in all molecules, and strong hydrogen bonding between a hydrogen atom and electronegative atoms like N, O, F. Hydrogen bonding gives ice its unusual properties and is important in biological molecules like DNA. The strengths of these different intermolecular forces influence physical properties of substances like melting and boiling points. Liquids exhibit surface tension, capillary action, and viscosity due to the attractive intermolecular forces between their molecules.
intermolecular_forces.ppt
Okay, let's break this down step-by-step:
* We are given: 1 mole of ice at -25°C
* Heat of fusion of ice = 6.01 kJ/mol
* Heat of vaporization of water = 40.7 kJ/mol
* Specific heat of ice = 2.09 J/g°C
* Specific heat of water = 4.18 J/g°C
1) Heat ice from -25°C to 0°C:
Q = m * c * ΔT
Q = (18 g) * (2.09 J/g°C) * (25°C) = 903 J
2) Heat
intern_forc_biceps2.ppt
Okay, let's break this down step-by-step:
* We are given: 1 mole of ice at -25°C
* Heat of fusion of ice = 6.01 kJ/mol
* Heat of vaporization of water = 40.7 kJ/mol
* Specific heat of ice = 2.09 J/g°C
* Specific heat of water = 4.18 J/g°C
1) Heat ice from -25°C to 0°C:
Q = m * c * ΔT
Q = (18 g) * (2.09 J/g°C) * (25°C) = 903 J
2) Heat
Similar to I need help with CH4- What is its intermolecular force- Question 4 of.docx (20)
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Hi!For hydrogen fluoride the predominant intermolecular force is .pdf
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Intermolecular forces consist of four types Dipo.pdf
Intermolecular forces consist of four types Dipo.pdf
Chem 2 - Intermolecular Forces & Phases of Matter II
Chem 2 - Intermolecular Forces & Phases of Matter II
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
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ID-D Samei- -14- Which ofthe following should sever be evaporwedon-hpl.docx
ID:D Samei, -14. Which ofthe following should sever be evaporwedon.hple Phosphoric acid b. Teluene c. Triethylene glyoel. d. Diethyl ether e. Aoaty halicy lis acid 15. Which of the following staments about the foemation of adipic acid thom cyclohenone is fal the krtone. b. A brown peecipitate of manganese dionide forms d. The final prodact is an alkene 16. A separatory funnel costains watcr, esher and ate other compound Which of the following is the compound most Wkely found is the water lyer? a Phenoil b. Sodiam benzoale 17, One step in the preparusios of cyelohevene froes cyclshexamol is tranaferring the dislate into a separatory fineel with saturated sodiem chloride solusion What is the parpose of this step? a. To renove water froms the distillane To lower the semperature of the solution d. To oxidize the cyclohesene 1& The neutral compound 1 s in a separatory funnel that also contains water and other How would you separate the 1,4-dimethoxyhenneme Daan out the water lnr, tes evaporate de ether. Add HCL thee basify the water leyer with NaOH and flter the product Drain out the ether layer, then filter the water lryer Add NaOil, thos ackdify the water layer with HCI and filher the product a. 19. How would the boiling point of water be effected ifsand was added so the water? . There would be a docrease. b. There wouild be no effect c There woeld be an increase d There would be both an increase and a decrease. 20. The final product from the process of crystalliration is isclated by a Graviy fitration b. Extraction c. Vacuum filtration. d Gas chromalographry
Solution
14 - (b)
Ether should never be directly kept on a hot plate because it can increase the temperature which is sufficient to start a fire.
15 - (d)
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I- Physical Properties- Color-- Clear Odor- Nail polish Yemover (corre.docx
I. Physical Properties: Color:- Clear Odor: Nail polish Yemover (corrected) Boiling Point (while distilling): ID(corrected) Refractive Index (optional) Density (optional): II. Solubility Test Results: Test performed pH of an aq. EtOH solution: Test Results Conclusions No soulaility oxpaic laver on TOP dist. H20 aq. HC OreanicLaver on TOP N Solu bilrt No Solubi lity ag. NaHCO conc. H2SO4 Afier you finally identify the Unknown, write a balanced equation for any one reaction that occurred in the solubility tests:
Solution
Pinacolone is a ketone, hence it does not react with water( non-polar), HCl, NaOH, aq NaHCO 3 (no acid group) et. However, it gave yellow solution with H 2 SO 4 . H 2 SO 4 being a conc. acid, it chars the compound. Another probable reason is the formation of enol species in acidic medium. Enols are slightly polar, and due to conjugation of OH lone pair with the double bond, the solution becomes coloured.
Me 3 CCOCH 3 + H + = Me 3 CC(OH)=CH 2
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I- H low do you know when to name something as an ionic compound- 2- H.docx
I. H low do you know when to name something as an ionic compound? 2. How do you know when to name something as a molecular compound? 3. How does nami ing an ionic compound differ from naming a molecular compound? 4. How do you know when a polyatomic ion is present? 5. Why do we include a Roman numeral in the name for compounds containing a transition metal and not for main group metals?
Solution
1. When we see that a metal (cation) is bonded to a non-metal (anion) we detect it to an ionic compound and name it so.
2. When two elements of comparable electronegativity combine to form a molecule it then becomes a covalent compound.
3. In naming ionic compounds we generally name the cation and anion separately. Molecular compounds either have trivial names or are named according to IUPAC rules.
4. Polyatomic ions are placed in brackets and so we can detect them.
5. Transition metals can have more than one oxidation states which normal metals dont. So, the roman numerals are included to indicate oxidation states of the transition metals.
.
I- Which type of star oscillates in brightness in a pattern in such a.docx
I. Which type of star oscillates in brightness in a pattern in such a way that we can measure its luminosity a. Black hole b. Main Sequence star c. Wolf-Rayet Star d. Cepheid Variable star What is Cephus? 3. To use a Cepheid Variable to measure the distance, what do you actually measure? a. Apparent brightness b. Absolute brightness c. The period of brightness change d. The period of color change 4. Cepheid Variables are relatively dim stars. a. Trae b. False . How far out does a Cepheid Variable work to measure distance a In the Milky Way galaxy b. 10 million ly c. 100 million ly d. 1 billion ly 6. Who first used Cepheid Variables to measure the distance to the Andromeda Galaxy a. Henrietta Swan Leavitt b. Hank Caldwell c. John Dobson d. Edwin Hubble 7. The faster a Cepheid Variable oscillates in brightness, the brighter the star is a. Trae b. False 8. A standard candle is . Which standard candle involves the rotation spood of spiral galaxies a. Cepheid Variables b. Scully-Fisher c. Tully-Fisher Relation d. Type la Superova 10. What would you use to measure the rotation speed of a galaxy a. A flux capacitor b. A spectrograph c. Radar d. Doppler Effect 11. Type Il supernova involve a. The death of huge stars b. The death of a regular sized main sequence star c. The death of two main sequence stars d. A white dwarf e. All of the above f None of the above g A and D h. A and c i. C and D
Solution
1. Cepheid variable star oscillates in brightness in a pattern in such a way that we can measure its luminosity.
.
I- Describe the different bonds that exist between the different macro.docx
I. Describe the different bonds that exist between the different macromolecules (Big 4). If there are variations
in bonding that change physical properties make sure to mention that as well.
Solution
Hi,
Among the big 4,
1. Carbohydrates have glycosidic linkages (C-O-C bond) between monosaccharides which form a dehydrated polysaccharide. Larger number of glycosidic linkage leads to a decrease in solubility of saccharide in water.
2. Proteins have peptide bonds which are formed via dehydration between amino acid chains (COOH and NH2 group form CO-NH). These bonds help amino acid units to form large structures (sheet/helix) of proteins. The presence of these bonds help in stability of proteins.
3. Lipids have mostly non-bonded interactions (and hydrogen bonds at polar head of the lipid) the interactions/H-bonds interplay in self aggregation and micellization of lipids.
4. Nucleic acids are composed of 5 carbon sugars mostly, with phosphate group attached to a CH2 group as C-O-P bond. The two COP bonds help forming long chain structures like DNA, apart from COP, C-N bond is also formed between nucleic acid and nitrogenous bases (A,T,U,G,C).
Thanks,
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I- Anole Company 25 POINTS Anole Company was incorporated as a new bus.docx
I.              Anole Company 25 POINTS
Anole Company was incorporated as a new business on January 1, 2015. The company is authorized to issue 20,000 shares of $5 par value common stock and 10,000 shares of 6%, $10 par value, cumulative, participating preferred stock. On January 1, 2015, the company issued 8,000 shares of common stock for $15 per share and 2,000 shares of preferred stock for $30 per share. Net income for the year ended December 31, 2015, was $375,000.
1.Refer to the information about Anole Company.
The amount of Anole’s total contributed capital at December 31, 2015, is
2.Refer to the information about Anole Company.
The number of Anole’s unissued shares of common stock at December 31, 2015, is
3.Refer to the information about Anole Company.
Anole’s total stockholders\' equity reported on the balance sheet at December 31, 2015, is
Solution
1. Contributed capital:
= 8,000×$15+2,000×$30
= $180,000
2.Unissued shares of common stock:
= 20,000-8,000
= 12,000
3. Stockholders equity in balance sheet:
= 8,000×$15+2,000×$30+$375,000
= $555,000
.
I) What are secondary bonding interactions (also called charge transfe.docx
I)
What are secondary bonding interactions (also called charge transfer interactions) and which elements are they most relevant to?
II)
Which halogen can be used to support central atoms in their highest oxidation states?
Solution
I)In a complex if one component is a good electron donor and the other a good electron acceptor, charge transfer interactions may occur. Several trasition metal complexes show charge transfer interactions. Electrons migrate from orbitals which has predominantly ligand character to orbitals having predominantly metal character or vice versa depending on energy of the orbitals.These can be identified by high intensity bands in their electromagnetic spectra.
II)Among halogens Iodine is a good candidate as it is a good donor as well as a good acceptor. For metals in high oxidation state electron density can be transferred from iodine to metal centre more effectively (due to low electron affinity and high polarisability).This stabilizes the complex.
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I need to write a soil chemistry paper using scientific articles and j.docx
I need to write a soil chemistry paper using scientific articles and journals as references. Can people suggest specific topics that have a lot of avaiable informetion on them that I could use to write a 7-10 page paper including some graphs and charts. Some ideas I have include nutrient pollution and eutrophication, root structure and soil formation, if anyone has any good topics with a high amount of information I would greatly appreciate it.
Solution
Answer:
There are a few topics which can be suggested. Please see below:
1. Characterisation of soil particles and speciation of compounds in the soil.
2. Behavior of contaminants such as heavy metals at the soil-water interface in the environment, and soil health.
3. Assessment of nitrogen and phosphorus cycle processes in agricultural soils with respect to enhancing plant nutrient uptake. Specific research topics include: improve fertilizer N recovery by crops; the functions and dynamics of soil organic matter; appropriate practices for management of soil organic matter.
4. Fate of nanoparticles in the soil environment
5. Micronutrient Management for Sustainable Agriculture and Environment
6. Examination of physical soil characteristics that contribute to soil fertility, soil quality, transport and exchange processes in soils. Research topics include: soil physical tests to determine soil health; examination of natural porous media; transport theories; the exchange processes in soils.
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I need help writing this program in MIPS MARS assembly language 2- Mak.docx
I need help writing this program in MIPS MARS assembly language
2. Make a function call to read an integer from user (user input is in seconds)
- display that number in days
- ... hours
- ... Minutes
- .... Seconds.
Solution
GUI with point-and-click control and integrated editor
Easily editable register and memory values, similar to a spreadsheet
Display values in hexadecimal or decimal
Command line mode for instructors to test and evaluate many programs easily
Floating point registers, coprocessor1 and coprocessor2. St andard tool: bit-level view and edit of 32-bit floating point registers (screenshot).
Variable-speed single-step execution
\"Tool\" utility for MIPS control of simulated devices. S tandard tool: Cache performance analysis tool (screenshot).
Single-step backwards
Recall from the beginning of this text the demonstration program duplicated below:
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I need help writing a python 3 function- I need to write a function th.docx
I need help writing a python 3 function.
I need to write a function that takes 3 inputs which are lists (first names, middle ini., and last names ).
It needs to take those names and put them together correctly and then print them in a column, 1 on top of another, and so on. they need to be printed in aphabetical order of the last names.
Also, they need to have a border around the whole column and the initials need to be aligned vertically. if there is no initial, a space can just take its place to keep things aligned.
I feel like being able to look at the code and go over it line by line until I understand it and can recreate it myself is the best way I learn. If someone could create this function for me, that would be unbelievably helpful.
Thanks SO much for the help and sorry for any grammar issues, I\'m writing this from my phone
Solution
def createNames(firstName, middleName, lastName): #creating a function to generate names from the given lists and give output of names in a list
allNames=[]
for i in range(len(firstName)):
name=lastName[i]+\" \"+middleName[i]+\" \"+firstName[i]
allNames.append(name)
allNames.sort() Â Â #sorting the names in alphabatical order of lastNames
return allNames
firstNames=[] Â Â #give here firstNames
middleNames=[]Â Â #give here middleNames
lastNames=[]Â Â #give here lastNames
namesList = createNames(firstNames, middleNames, lastNames)
print(\"_______________________________\ \")
for name in namesList:
print(name+\"\ \")Â Â #printing the names one by one
print(\"_______________________________\")
.
I need help with my geotechnical engineering HW-- Can a soil have a lo.docx
I need help with my geotechnical engineering HW,,
Can a soil have a low void ratio and a low total unit weight at the same time? Explain. Name and define three types of weathering. From memory, try to write the equations for water content, void ratio, dry unit weight, moist (total) unit weight, and saturated unit weight. After 5 minutes you can look up the definitions, and write the correct answers below your attempt. For a given soil, the following parameters are known: e -0.62, specific gravity 2.65, moisture content w=15% Determine: 3. 4. 5. 6. dry unit weight moist unit weight water content at S-100% saturated unit weight .
Solution
3. No, low void ratio and low total unit weight at the same time are possible, beacause if void ratio is low ultimately the mass of the solids is high so that the unit weight increases.
4.Types of Weathering:
i) Mechanical weathering: which produces clasts of rock/mineral which are smaller that the original and of the same general composition. There are two basic processes: Fracturing and Abrasion
ii) Chemical weathering: It also produces smaller pieces, but of a different chemical composition, the minerals have been changed in an attempt to be more in balance with surface conditions.
iii) Biological weathering: It is the action of plants and animals can and do lead to both mechanical and biological weathering.
5. Water Content = Weight of water/weight of solids
Void ratio = volume of voids/volume of solids
Moist unit weight = mass of soil * accleration due to gravity / volume of soil
Dry unit weight = mass of solid * accleration due to gravity / volume of soil or moist unit wieght/(1+water content)
Saturated unit weight = submerged unit wiegt + unit weight of water
6. Dry unit Weight = 16.35 KN/cubic meter
Moist unit Weight = 18.81 KN/cubic meter
Water content at S=100%, w=23.39%
Saturated Unit Weight = 20.185 KN/cubic meter
.
I need help m5 In the modern Respond to the following questions with s.docx
I need help
m5
In the modern Respond to the following questions with supporting details from what you\'ve learn 1. W 2. Consider your social media relationships. Does texting, tweeting, and other social media strengthen our relationships with world, there are so many gadgets that keep us \"connected.\" But what are we really connecting to? ed in this module: hat are some of the pros and cons to the widespread acceptance of communication through technology? others, or does it create fake relationships? 3. Do the friends you have online meet your needs? 4. If you do not use social media, discuss your decision to avoid these innovations. 6. What has your experience been like in keeping, maintaining, and growing relationships with others who do use them? 5. Why have you decided to not partake in these technologies? 7. What do you imagine are the costs and benefits to the next generation who will grow up with all of these technologies already integrated into their culture?
Solution
5th part:
The decision to not partske in these technologies may be because of following reasons:
1. These technologies keep people engaged and people tend to neglect the current outgoing activities to their surroundings. They tend to ignore there most neighboring surroundings to connect with a distant friend.
2. These technologies are proved to be harmful for health when used excessively. People tend to get used to these and once these go into habit, they use it excessively.
3. These make distractions from work or study and people lose their concentration becaus eof a random text or call.
4. These make people inactive in life as more social media less sports
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I need help describing the two kinds of people who are responsible for.docx
I need help describing the two kinds of people who are responsible for building and securing the network cryptography and the people who try to break it.
Solution
Network cryptography:
Cyber security is the most significant field in the IT sector and it try to reduce the risk of hacking.there are many people with different positions to preserve security.
There are many other people who take care of cybersecurity in the IT field.
Security analyst
Security engineer
Security architect
Security administrator
1)Analyzing and assessing various vulnerabilities in the software and hardware .and then think of what are the existing and available tools and procedures to eradicate the detected vulnerabilities, and finally suggests the best outputs.
2) Inspect the security issues and then give responses immediately to reduce the vulnarabilities.
3) He is the person who is involving in the design of the system security
4) Who manages all the system securities?
on the other hand is the hacker
Hacker: In the context of security a hacker is the one who trying to get the information from a compuer network.
They may have different reasons:
Because of profits, interest, to exhibit their talent etc..
.
12- Your friend has high protein and sugar in her urine- She repeats t.docx
12. Your friend has high protein and sugar in her urine. She repeats the tests and determines that these components are repeatable, suggesting they are present in her urine. What does this signify to you, as her friend and primary care physician?
Solution
Everyone has protein in their blood. The main protein in your blood is called albumin. healthy kidneys should only filter tiny (trace) amounts into the urine as most protein molecules are too large for the filters (glomeruli). When she have protein in her urine, it is called proteinuria (or albuminuria), which is usually a sign of chronic kidney disease. As proteinuria progresses, oedema may occur. High levels of glucose in her urine may indicate hyperglycemia, or high blood sugar, which often indicates the onset of diabetes. Untreated hyperglycemia can lead to ketoacidosis, which can result in diabetic coma and even death.
Possible causes of proteinuria:
1. Diseases of the glomeruli like glomerulonephritis or diabetes.
2. Urine infection such as cystitis/urinary tract infections.
3. Congestive heart failure, a first warning of eclampsia in pregnancy.
4. May be she tested concentrated urine samples (early morning), or if she have become dehydrated or haven’t drunk much. It is also better to check protein/creatinine ratio (PCR test).
5. Temporary proteinuria may occur after vigorous exercise or if she have a high fever.
Hope this helped you!
Thank You So Much! Please Rate this answer as you wish.(\"Thumbs Up\")
.
12- Under which of the following circumstances is the principal-agent.docx
12. Under which of the following circumstances is the principal-agent problem likely to be most serious.
Between general practitioners and patients.
Between surgeons and patients.
Between hospitals and nurses.
Between dentists and physicians.
Between physicians and lawyers.
13. Inoculation programs against certain diseases such as small pox, polio, and whooping cough create
public goods.
positive externalities in consumption.
nonrival goods.
nonexcludable goods.
external costs to society equal to the costs of the program.
14. The government agency that controls the development and availability of new medical discoveries, including pharmaceutical drugs is
the Health Care Financing Administration.
the Food and Drug Administration.
the Centers for Disease Control.
the National Institutes for Health.
the Federal Emergency Medical Administration.
Solution
12) Between general practitioners and patients.
13) positive externalities in consumption.
14) the Food and Drug Administration
.
12- An Asteroid passes by Earth every 17 years- The asteroid orbits th.docx
12. An Asteroid passes by Earth every 17 years. The asteroid orbits the Sun, not the Earth. Remember these: M(Earth) 5.97 x 1024kg, R(Earth) 6400 km, M(Sun)- 1.99 x 1030kg, R(Sun) 700, 000 km, and the distance between the Earth and the Sun is 1.5 x 1011m. Answer the following: What is the radius of its orbit? What is the closest it gets to Earth? What is its angular velocity?
Solution
Time period of the Asteroid, T = 17 years
= 17*365*24*60*60
= 5.361*10^8 s
let r is the radius of the orbit of the asteroid.
we know, T = 2*pi*r^1.5/sqrt(G*Msun)
T^2 = 4*pi^2*r^3/(G*M_sun)
r^3 = G*M_sun*T^2/4*pi^2
r = (G*M_sun*T^2/4*pi^2)^(1/3)
= (6.67*10^-11*1.99*10^30*(5.361*10^8)^2/(4*pi^2))^(1/3)
= 9.886*10^11 m <<<<<<<---------Answer
distance between the earth and the asteroid = 9.886*10^11 - 1.5*10^11
8.386*10^11 m <<<<<<<---------Answer
Angular velocity, w = 2*pi/T
= 2*pi/(5.361*10^8)
= 1.172*10^-8 rad/s <<<<<<<---------Answer
.
11- What are the components of the user interface-SolutionNow a days t.docx
11. What are the components of the user interface?
Solution
Now a days there are lots of types of user interfaces. We are going to describe the components in general, using four main categories:
Containers: This are panes, windows, forms which hold other components. Containers have grids in order to arrange other components. There are rows and columns grids, flexible grids, flow grids. Containers also has the responsibility to adapt to the screen size. Some containers are:
Window
Frame
Form
Dialog
Menus: This user interface components allows the user to navigate through the different commands of a given applications. Menus organize commands in a logical and hierarchical manner, using several submenus. Some types of menu are:
Main Manu (located on the top of the container).
Menu Bar
Contextual Menu
Controls. This components allows communication between the user and the application. This allows the user to send and receive request, messages and information from/to the application. This the category which hold the bigger number of components. Some of them are the following:
Text Boxes
Text Area
Buttons
File Control
Select List
Check
Optinon
Label
Tables
Graphic Area
Media
Controllers: The controller is not a graphic component but it is necessary to process input/output. This are pieces of code which capture user interaction, takes arguments and details and finally process commands and requests. Normally the controller uses the publisher/subscriber pattern. It subscribes itself with the components and receive notifications when the events occur, being able to react accordingly.
.
11- The employer payroll tax expense on employee wages that is paid fo.docx
11. The employer payroll tax expense on employee wages that is paid for federal funding of state compensation for unemployed workers is:
Social Security and Medicare tax expense
Federal Unemployment Tax (FUTA) expense
State Unemployment Tax (SUTA) expense
Worker\'s Compensation expense
Retirement plan voluntary deductions
Health insurance voluntary deductions
Contributions
None of the above
Solution
11)Â Â Federal Unemployment Tax (FUTA) expense
12) click on the Pay Employees icon and select Set Up Payroll Service
13) All of the above
14) They are deducted from the employee\'s gross pay
15) General tab
16) Medicare tax withholding
17) Maintain Employees & Sales Reps
18) Employee Icon
19) Pay info tab
20) None of the above
.
110) When you create a form from two tables that have a one-to-many re.docx
110) When you create a form from two tables that have a one-to-many relationship, the first table selected becomes the subform and the second table you select becomes the main form. teue or false
Solution
.
12) )Compute the July 2014 cost of capital (rounded to nearest percent.docx
12) )Compute the July 2014 cost of capital (rounded to nearest percent) for an investment center with the following information:
Pre-tax operating income for July 2014 $15,500,000
Assets at July 31, 2014 $8,200,000
Current liabilities at July 31, 2014 $3,200,000
Long-term liabilities at July 31, 2014 $2,200,000
Income tax expense for July 31, 2014 $5,500,000
EVA $9,500,000
Solution
Stockholder\'s Equity = Assets - Current Liabilities - Lon Term Liabilities
Stockholder\'s Equity = 8,200,000 - 3,200,000 - 2,200,000 = $2,800,000
After Tax Operating Income = 15,500,000 - 5,500,000 = $10,000,000
EVA = After Tax Operating Income - Cost of Capital
Cost of Capital = 10,000,000 - 9,500,000 = $500,000
Cost of Capital in % = 500,000 / 2,800,000 = 17.86%
.
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