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Sergeeva-Olga [200]
3 years ago
11

In addition to intermolecular forces affecting liquid properties, they also influence gas properties but to a smaller degree. On

e property that is influenced for a gas is the pressure. Intermolecular forces make the molecules attracted to one another which will affect the pressure that they exert on the container walls. This is one of the reasons why not all real gases act "ideally" and why the ideal gas law is not 100% accurate. Consider that you have one rigid container filled with 1 mole of O2 gas and another rigid container filled with 1 mole of Cl2 gas, both of which have the same volume and temperature. Which gas is expected to have a higher pressure and why?
a. O2, because it has stronger intermolecular forces
b. O2, because it has weaker intermolecular forces
c. Cl2, because it has stronger intermolecular forces
d. Cl2, because it has weaker intermolecular forces
Chemistry
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

b) O2, because it has weaker intermolecular forces

Explanation:

The preassure is produced by the collisions of the gas molecules with the walls of its container.

When the intermolecular forces between the gas molecules increase, those molecules start to "slow down" by effect of the interactions. The collisions decrease in frequency and intensity producing a smaller preassure in the container.

Both O2 and Cl2 are non-polar gases and the only intermolecular forces they have are the London ones. Given that the O2 molecules are smaller than the Cl2, the last ones attract each other with more strengh.

Being all that said, the container with the oxygen is expected to have a higher preassure.

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In a 1.0× 10–6 M solution of Ba(OH)2(aq) at 25 °C, identify the relative molar amounts of these species.
Marysya12 [62]
Thank you for posting your question here. Below is the solution:

HNO3 --> H+ + NO3- 
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>

<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>


<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
4 0
3 years ago
Doping Se with P would produce a(n) ________ semiconductor with ____________ conductivity compared to pure Se.
deff fn [24]
Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se

the reason is P dopant will introduce holes in the Se as P has lesser valence electron
4 0
3 years ago
Define Saturated Solution. What happens as a saturates solution cools?
solong [7]

Saturated solution is a solution in which no more solute can be dissolved in the solvent. When saturated solution cools, the solution began precipitate from the solution, because under lower temperature, usually, less amount solute can be dissolved in the solvent.

8 0
3 years ago
Read 2 more answers
What is the percent composition of nitrogen in N 2 O
mash [69]

Answer:

63.6%

Explanation:

The given compound is:

     N₂O;

The problem here is to find the percent composition of nitrogen in the compound.

First find the molar mass of the compound:

 Molar mass of N₂O = 2(14) + 16  = 44g/mol

So;

 Percentage composition of Nitrogen  = \frac{2 x 14}{44}  x 100  = 63.6%

5 0
3 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
3 years ago
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