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laila [671]
3 years ago
13

The electron cloud of HF is smaller than that of F2, however, HF has a much higher boiling point than F2 has. Which of the follo

wing explains how the dispersion-force model of intermolecular attraction does not account for the unusually high boiling point of HF?
A. F2 is soluble in water, whereas HF is insoluble in water.
B. The F2 molecule has a greater mass than the HF molecule has.
C. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between H+ and F- ions.
D. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has both weak dispersion force attractions and hydrogen bonding interactions between its molecules.
Chemistry
1 answer:
mafiozo [28]3 years ago
7 0

Answer:D

Explanation:

The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.

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What is the oxidation number for s in s2o4?
hoa [83]
The oxidation number of O is -2 and there's 4 of them so the total is -8. Since it's a molecule the sum of the oxidation numbers of the atoms must equal 0. If 4 oxygen atoms have an oxidation number of -8 then the two S atoms must is +8 to equal 0 so the oxidation number of the S is 8/2 = +4
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3 years ago
Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th
ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}      .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol

The chemical equation for the reaction of ammonia and HCl follows:

NH_3+HCl\rightarrow NH_4^++Cl^-

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonia

  • Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonium ion

  • Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.8\times 10^{-5}

10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}

The chemical equation for the dissociation of ammonium ion follows:

NH_4^+\rightarrow NH_3+H^+

The expression of K_a for above equation follows:

K_a=\frac{[NH_3][H^+]}{[NH_4^+]}

We know that:

[NH_3]=[H^+]=x

[NH_4^+]=0.072M

Putting values in above expression, we get:

5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M

To calculate the pH concentration, we use the equation:

pH=-\log[H^+]

We are given:

[H^+]=6.32\times 10^{--6}M

pH=-\log (6.32\times 10^{-6})\\\\pH=5.20

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0
3 years ago
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statuscvo [17]

Answer:

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Explanation:

8 0
3 years ago
The mass spectrum of an unknown compound has a molecular ion peak with a relative intensity of 57.10% and an M+1 peak of 6.83%.
max2010maxim [7]

There are 11 Carbon atoms in the compound.

<u>Solution:</u>

Carbon atom count is the ratio of the M peak to the M+1 peak.

\text{ Number of Carbon atoms }=\frac{\text { Relative intensity of } M+1 \text { peak }}{0.011 \times \text { Relative tntensity of } M \text { peak }}

Here M peak is 57.10% and M+1 peak is 6.83%. On applying the values in the formula we get,

\frac{0.0683}{0.011\times0.571g} = 10.87\approx11

Therefore, the number of Carbon atoms in the compound are 11.

Refer the image attached below for a better understanding of M peak and M+1 peak.

The heaviest ion that has the greatest m/z value is said to be the molecular ion peak in mass spectrum.

6 0
4 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

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Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

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Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

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Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

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Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
4 years ago
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