which of the following molecules would you expect to have a dipole moment of zero? a,CH2 Ch3
bH2C=0
cCH2cl
dNH3​

Answers

Answer 1

Answer: The molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.

Explanation:

The product of magnitude of the charge calculated in electrostatic units is called dipole moment.

Formula for dipole moment is as follows.

Dipole moment = Charge (in esu) [tex]\times[/tex] distance (in cm)

Non-polar molecules have zero dipole moment.

For example, [tex]CH_{3}-CH_{3}[/tex] is a non-polar molecule so its dipole moment is zero.

[tex]H_{2}C=O[/tex] is a polar molecule so it will have dipole moment.

[tex]CH_{2}Cl_{2}[/tex] is a polar molecule so it will have dipole moment.

[tex]NH_{3}[/tex] has nitrogen atom as more electronegative than hydrogen atom. So, net dipole moment will be in the direction of nitrogen atom.

Thus, we can conclude that the molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.


Related Questions

Match the name of each gas law to the properties it compares. (3 points)
1. Boyle's law
2. Avogadro's law
3. Gay-Lussac's law
a. Volume and moles
b. Pressure and volume
c. Pressure and temperature

Answers

Answer:

1. B

2. A

3. C

Explanation:

1. Boyle's law is one of the gas laws that states that the pressure of a gas is inversely proportional to its volume at a constant temperature. PV = K. Hence, this gas law compares the properties of the pressure (P) and the volume (V)

2. Avogadro's law states that the volume of a gas is directly proportional to the number of molecules of that gas, at a constant temperature and pressure. K = Vn. Hence, this gas law compares the properties of the number of moles (n) and the volume (V).

3. Gay-Lussac's law states that the pressure of a gas is directly proportional with the temperature of the gas at a constant volume. K = PT. Hence, this law compares the properties of the pressure (P) and the temperature (T)

The properties compared by Boyle's law are pressure and volume, Avogadro's law is volume and moles, and Gay-Lussac's law is pressure and temperature.

What are the variations of ideal gas law?

The ideal gas has the absence of interatomic collisions and follows the ideal gas equation.

The ideal gas law at constant pressure and temperature is termed Boyle's and Gay Lussac's law. The laws can be given as:

Boyle's law was the nature of ideal gas at a constant temperature and compares the pressure and volume of the gas.

Avogadro's law was the ideal nature of gas, with the moles of gas and volume in specific value at constant temperature and pressure.

Gay-Lussac's law compares the pressure and temperature of the gas at constant volume.

Learn  more about ideal gas laws, here:

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What is the Ke of the 20 gram shoe as it falls to the ground at 6 m/s?


PLEASE HELP ITS 7TH GRADE SCIENCE!!!

Answers

It’s 0.36 J
Bc I did this question already and I think I got it right I’m very sorry if you get it wrong

Which statement shows why mass is conserved for this reaction? 2Mg + O2 → 2MgO

Answers

Answer:

The same number of each atom are on both sides of the equation.

The Law of conservation of mass states that matter is neither destroyed nor created in a chemical reaction. Therefore, the mass of the reactants will be equal to the mass of the products in a chemical reaction. According to the law matter is neither created nor destroyed in a chemical reaction.

An ether and alkene are formed as by products in this reaction. draw the structures of these by-products and give mechanisms for their formation

Answers

Answer:

sim eu também preciso desta respota

An ether and alkene are formed as by products in the reaction which is a electrophilic addition reaction.

What is electrophilic addition reaction?

An addition reaction known as an electrophilic addition reaction occurs when a chemical molecule having a double or triple bond has one of its bonds broken and two new bonds are formed. The interconversion of C=C and CC into a variety of significant functional groups, such as alkyl halides and alcohols, is made possible via the key.

The following describes the general mechanism: Hydrogen bromide produces an electrophile, H+, which attacks the double bond to create a carbocation. The production of ions is dominated by secondary carbocation because it is more stable than primary carbocation.

Thus, an ether and alkene are formed as by products in the reaction which is a electrophilic addition reaction.

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what is a property of every mixture

Answers

Explanation:

can u post a picture of the question ?

g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM solution and ___________ compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Answers

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing point of pure . Calculate the mass of potassium bromide that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for potassium bromide in .

Answers

The question is incomplete, the complete question is:

When 177. g of alanine [tex](C_3H_7NO_2)[/tex] are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is [tex]5.9^oC[/tex] lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is [tex]7.2^oC[/tex] lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\Delta T_f=i\times K_f\times m[/tex]

OR

[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

When alanine is dissolved in mystery liquid X:

[tex]\Delta T_f=5.9^oC[/tex]

i = Vant Hoff factor = 1 (for non-electrolytes)

[tex]K_f[/tex] = freezing point depression constant

[tex]m_{solute}[/tex] = Given mass of solute (alanine) = 177. g

[tex]M_{solute}[/tex] = Molar mass of solute (alanine) = 89 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

[tex]5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m[/tex]

When KBr is dissolved in mystery liquid X:

[tex]\Delta T_f=7.2^oC[/tex]

i = Vant Hoff factor = ?

[tex]K_f[/tex] = freezing point depression constant = [tex]2.37^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute (KBr) = 177. g

[tex]M_{solute}[/tex] = Molar mass of solute (KBr) = 119 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

[tex]7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63[/tex]

Hence, the van't Hoff factor for potassium bromide in X is 1.63

In Denver, Colorado the elevation is about 5,280 feet above sea level. Explain what potential effects this may have on the solubility of a gaseous solute in a liquid solution.

Answers

Answer:

The solubility of the gaseous solute decreases

Explanation:

As we know, pressure decreases with altitude. This means that, at higher altitudes, the pressure is much lower than it is at sea level.

The solubility of a gas increases with increase in pressure and decreases with decrease in pressure.

Hence, in Denver, Colorado where the elevation is about 5,280 feet above sea level, a gaseous solute is less soluble than it is at sea level due to the lower pressure at such high altitude.

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.
pH= 2.89

Answers

Answer: The value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].

Explanation:

pH is the negative logarithm of concentration of hydrogen ion.

It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.

[tex]pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M[/tex]

The relation between pH and pOH value is as follows.

pH + pOH = 14

0.0012 + pOH = 14

pOH = 14 - 0.0012 = 13.99

Now, pOH is the negative logarithm of concentration of hydroxide ions.

Hence, [tex][OH^{-}][/tex] is calculated as follows.

[tex]pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M[/tex]

Thus, we can conclude that the value of [tex][H_{3}O^{+}][/tex] is 0.0012 M and [tex][OH^{-}][/tex] is [tex]1.02 \times 10^{-14}[/tex].

Give the IUPAC name for the following alkyl group, and classify it as primary, secondary, or tertiary.

CH3(CH2)9CH2

Answers

Answer:

Give the IUPAC name for the following alkyl group, and classify it as primary, secondary, or tertiary.

CH3(CH2)9CH2

Explanation:

In the given alkyl group there are eleven carbon atoms.

So, the alkyl group name is:

n-undecyl.

Pimary carbon is the one which is attached only one other carbon atom,group.

Secondary carbon is the one which is attached to two carbons.

Thertiary carbon is the one which is attached to three other carbons.

In the given alkyl group,

the primary,secondary alkyl groups are shown below:

There is no tertiary carbon atom in the given molecule.

Calculate the concentration of a solution with 0.8g of NaCl in 280mL of water.

Answers

Answer: The molarity of NaCl solution is 0.0489 M

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)

We are given:

Given mass of NaCl = 0.8 g

Molar mass of NaCl = 58.44 g/mol

Volume of the solution = 280 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of solution}=\frac{0.8\times 1000}{58.44\times 280}\\\\\text{Molarity of solution}=0.0489M[/tex]

Hence, the molarity of NaCl solution is 0.0489 M

help please and thank you!

Answers

Answer:

a) N2(g) + H20 (aq) --> HNO3 (aq) + NO (g)b) 40 NO2(g) + H20(aq) = 20 HNO3(aq) + 20 NO(g)

Explanation:

Draw the Lewis structure of NCl3NCl3 . Include lone pairs. Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Select Draw Rings More Erase Cl N

Answers

Answer:

See explanation and image attached

Explanation:

Nitrogen trichloride NCl3 contains one nitrogen and three chlorine atoms. Each chlorine atom has three lone pairs of electrons while nitrogen has one lone pair of electrons.

The molecule geometry of the molecule is trigonal pyramidal while it's electron domain geometry is tetrahedral. Nitrogen is the central atom and is found in sp3 hybridization.

There is no formal charge on the molecule.

table salt is 42.7% sodium how many grams of salt contain 76 g of sodium

Answers

Answer:

Số gam muối ăn cần là 76:42,7%=177,986 g

Explanation:

phương pháp VI PHÂN ĐỒ THỊ để xác định bậc phản ứng

Answers

Answer:

mwlooy kagabi jal

64 JAHA VI PHÂN KAY

Which of the following is an example of a nonrenewable resource?
a
cattle
b
uranium
c
cotton
d
trees

Answers

Answer:

b . uranium, It is not a renewable resource.

Can the properties of different substances in a mixture be used to separate them?

Answers

Answer: here you go

Explanation:

Physical properties of the substances in a mixture are different, so this allows the substances to be separated. Think about the example of a mixture of salt water.

Answer:

Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions, and colloids. The components of a mixture retain their own physical properties. These properties can be used to separate the components by filtering, boiling, or other physical processes.

Explanation:

..............................................................................................

Answers

………………………………………………………………………………..

The seagulls on the beach -​

Answers

Seagulls like beaches because they can find food

điện phân nóng chảy hoàn toàn 14,9 gam muối clorua của 1 kim loại kềm R thu đc 2,24 lít khí . R là kim loại gì ?

Answers

Answer:

d

Explanation:

nCl2 = 0,1 -> nRCI = 0,2

M muéi = R + 35,5 = 14,9/0,2

-> R = 39: R la K

Answer:

d

Explanation:

nCl2 = 0,1 -> nRCI = 0,2

M muéi = R + 35,5 = 14,9 / 0,2

-> R = 39: R la K

g Ions B and C react to form the complex BC. If 35.0 mL of 1.00 M B is combined with 35.0 mL of 1.00 M C, 0.00500 mol of BC is formed. Determine the equilibrium constant for this reaction.

Answers

Answer:

Kf = 0.389.

Explanation:

Hello there!

In this case, it turns out possible for us to solve this problem by firstly writing the equilibrium chemical equation and equilibrium expression for the formation of this complex:

[tex]B+C\rightleftharpoons BC\\\\Kf=\frac{[BC]}{[B][C]}[/tex]

Thus, we firstly calculate the concentrations at equilibrium, knowing that the reaction extent in this case is 0.00500mol (same as the formed moles of BC):

[tex][B]=[C]=\frac{0.0350L*1.00mol/L-0.00500mol}{0.0700L} =0.429M[/tex]

[tex][BC]=\frac{0.00500mol}{0.0700L} =0.0714M[/tex]

And finally, the equilibrium constant:

[tex]Kf=\frac{0.0714}{[0.429][0.429]}\\\\Kf=0.389[/tex]

Regards!

Select all the correct locations on the image.
Select the areas that would receive snowfall because of the lake effect.
(The right answer for this question, I got it right on the edmentrum)

Answers

Answer:

A:Itasca

A:ItascaB:Hubbard

A:ItascaB:HubbardC:Douglas

A:ItascaB:HubbardC:DouglasD:Grand Marais

E:Two harbors

F:Duluth

Answer:

Duluth, Twin Harbors and Grand Marais because they are on the coast of the lake.

Explanation:

What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?

Answers

Answer:

0.2 M

Explanation:

Step 1: Given data

Mass of sugar (sucrose): 15 gVolume of water: 0.2 L (we will assume it is the volume of the solution)

There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.

Step 2: Calculate the moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

15 g × 1 mol/342.3 g = 0.044 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.044 mol/0.2 L = 0.2 M

The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains after 28 seconds? This reaction was found to be first order.

Answers

Answer: The mass of sample that remained is 0.127 mg

Explanation:

The integrated rate law equation for first-order kinetics:

[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(1)

Given values:

a = initial concentration of reactant = 0.500 mg

a - x = concentration of reactant left after time 't' = ?mg

t = time period = 28 s

k = rate constant = [tex]0.049s^{-1}[/tex]

Putting values in equation 1:

[tex]0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg[/tex]

Hence, the mass of sample that remained is 0.127 mg

The hydrogen fluoride molecule, HF, is more polar than a water molecule, H2O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.

Answers

Answer:

Water forms more hydrogen bonds than HF

Explanation:

The answer to this question goes back to the idea of hydrogen bonding. Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom such as fluorine or oxygen.

However, in HF, there are three lone pairs of electrons on fluorine atom and one hydrogen atom bonded to fluorine.

In H2O, there are two lone pairs of electrons on oxygen atom and two hydrogen atoms bonded to oxygen. This simply means that water can form four hydrogen bonds while HF only forms two hydrogen bonds.

This implies that H2O molecules possess more hydrogen bonding than HF molecules. Hence, the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water.

5 compounds that has electrovalent and covalent bond

Answers

Answer:

electrovalent

NaCl

Lithium Carbonate

ammonium phosphate

aluminium floride

potassium hydride

covalent

methane

benzene

carbon iv oxide

hydro flouride

hydro chloride

the mixture of base and acid

Answers

Answer:

Mixture of a Strong Acid and a Strong Base

On mixing a strong acid and strong base neutralization (pH = 7) takes place. The resulting solution may be an acid or base depending on the Concentration. Say, N1, V1 is the strength and volume of the strong acid and N2, V2 is the strength and volume of the strong base

Explanation:

Most introductory chemistry books will teach that the reaction between an acid and a base is called neutralization, and the products formed are water and a salt

Which of the following is NOT likely to cause a change in average annual temperatures on Earth?
​a. Human activity. b. Solar eclipses.
​c. Photosynthesis by plants and algae. ​d. Strength of solar radiation.

Answers

Answer:

i think C  is the answer

Explanation:

The change in average annual temperatures on earth will be due to "photosynthesis by plants and algae".

What is photosynthesis?

Photosynthesis can be defined as a process in which plants, as well as other organisms, as well as other organisms, utilize to transform sunlight into chemical energy which can then be released to power the organism's activities using cellular respiration.

What is plants?

Plants seem to be mostly photosynthetic eukaryotes belonging to the plantae kingdom.

Therefore, photosynthesis cannot change in average annual temperature on Earth.

To know more about photosynthesis.

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convert 12nanometer to centimeter

Answers

Answer:

1x10^-6

Explanation:

What is the pCu of the resulting solution if 20.00 mL of 0.08 M EDTA (H4Y) is added to 15.00 mL of 0.10 M CuSO4 and buffered at pH 10? The Kf’ for complex CuY2- is 2.21 x 1018

Answers

Answer:

The answer is "5.4".

Explanation:

[tex]BoH + HCL =BCL +H_2o \\\\At eq \\\\N_1V_1=N_2V_2 \\\\v_2=20 \ ml\\\\[BCL]=\frac{20 \times 0.08}{20+20}=0.04\\\\pH = \frac{1}{2} [pkw - pk_b - \log e]\\\\pk_b = 2 pH - Pkw + \Log C\\\\pK_b=5.4[/tex]

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