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Dominik [7]
2 years ago
10

What type of van der Waals interactions occur between molecules

Chemistry
1 answer:
BigorU [14]2 years ago
7 0

Given what we know, we can confirm that the type of van der Waals interactions that occur between molecules of O2, SCl2, and CH4 in liquids of these substances are the presence of <u>London dispersion forces</u>.

<h3>What are London dispersion forces?</h3>
  • They are a force of attraction between atoms.
  • They are generated by electrostatic attraction.
  • These forces are common between atoms in close proximity and occur often when compounds have a symmetrical distribution of atoms.
  • They are generated by the formation of temporary dipoles.

Therefore, given the symmetry of the atoms disposition in these compounds and the temporary dipoles generated by the atoms being in close proximity, we can confirm that the van der Waals forces present in each compound are London dispersion forces.

To learn more about van deer Waals forces visit:
brainly.com/question/13201335?referrer=searchResults

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Calculate the amount of heat required for the evaporation of 105.0 g of water at 100.0 c
son4ous [18]
To determine the amount of heat needed to evaporate a certain amount of water at its boiling point, we use the equation,

                  H = m(Hv)

where H is the heat, m is the mass (in grams), and Hv is the latent heat of vaporization. Substituting the known values,

                 H = (105 g)(540 cal/g)

                  H = 56700 cal

                H = 56.7 kcal

Answer: 56.7 kcal
8 0
4 years ago
(?)Na2HPO4 → (?)Na4P2O7 + (?)H2O
Akimi4 [234]

Answer: 2,1,1

Explanation: (2)Na2HPO4=(1)Na4P2O7+(1)H20

7 0
4 years ago
If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79
Helen [10]
1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass. 

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4

3) convert moles of CuSO₄ to moles of Cu

1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu

I did it step-by-step as the explanation but you can do all of this in one step. 

200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu


4 0
4 years ago
A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi
viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette.  A blank had an absorbance of 0.029.  The absorbance of an unknown solution of compound A was 0.375.  Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = 1502.53\ {Ms}^{-1}

Explanation:

According to the Lambert's Beer law:-

A=\epsilon l c

Where, A is the absorbance

 l is the path length  

\epsilon is the molar absorptivity

c is the concentration.  

Given that:-

c = 3.96\times 10^{-4}\ M

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M

\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}

<u>Molar absorptivity of compound A = 1502.53\ {Ms}^{-1}</u>

4 0
4 years ago
A student is given an antacid tablet that weighed 5.6832 g. The tablet was crushed and 4.3628 g of he antacid was added to 200 m
SSSSS [86.1K]

Answer:

a) 8.33 ml of the original stomach acid is neutralized

b) 191.67 ml of the stomach acid was neutralized

c)  249.68 ml acid would be neutralized by the original tablet

Explanation:

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid

8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid

b) how much stomach acid was neutralized y the 4.3628 g tablet?

It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)

This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml

c) how much stomach acid would have been neutralized by the original 5.6832 g tablet

4.3628 g antacid is equivalent to 191.67 ml acid  ( this is the answer for question 2)

5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid

3 0
3 years ago
Read 2 more answers
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