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leva [86]
3 years ago
9

Sort each of the following events based on whether the solubility of the indicated gas will increase, decrease, or stay the same

. Each phrase specifies the gas involved and the change in its environment.
Put the following in one of three categories.
"gas solubility increases" Gas solubility decreases" " gas solubility does not change"

1. the pressure of a gas over a solvent is increase
2. the partial pressure of an anesthetic gas is increased
3. air in blood a diver descends 10 M and pressure increases by 1 atm
4. the temp is increase
5. O2 the temp of a body of water rises.
Chemistry
2 answers:
lbvjy [14]3 years ago
7 0

1. Gas solubility increases if the pressure of the gas over the solvent is increased.

2. Gas solubility increases if the partial pressure of the anesthetic gas is increased.

3. Gas solubility increases if a diver descends 10 m and pressure is increased by 1 atm.

4. Gas solubility decreases if the temperature is increased.

5. Gas solubility decreases if the temperature of the body rises.

Further Explanation:

Solubility:

It is the chemical property by virtue of which any substance is capable to dissolve in other substances. It is measured in terms of the maximum amount of solute that can be dissolved in the given amount of solvent. It is dependent on intermolecular forces, temperature, and pressure

There are weak intermolecular forces between the gas molecules. The average kinetic energy increases with the increase in temperature. This is because the molecules start vibrating about their mean positions. As a result, weak forces are overcome and the molecules re-enter into the gas phase. So solubility decreases with the increase in temperature.

Henry’s law depicts the relationship of gas solubility with its partial pressure. It states that the solubility is directly proportional to the partial pressure of the gas. Higher the partial pressure of the gas, more will be its solubility and vice-versa.

The expression for Henry’s law is,

 {{\text{S}}_{{\text{gas}}}}={{\text{k}}_{\text{H}}}\cdot {{\text{P}}_{{\text{gas}}}}

Here,

{{\text{S}}_{{\text{gas}}}} is the solubility of the gas.

{{\text{k}}_{\text{H}}} is Henry’s constant.

{{\text{P}}_{{\text{gas}}}} is the pressure of the gas.

1. The solubility of gas is directly proportional to the partial pressure of the gas in accordance with Henry’s Law. So the solubility of gas increases with the increase in its partial pressure.

2. The solubility of gas is directly proportional to the partial pressure of the gas in accordance with Henry’s Law. So the solubility of anesthetic gas increases with the increase in its partial pressure.

3. As the diver descends 10 m and the pressure is increased by 1 atm, solubility of gas increases in accordance with Henry’s Law.

4. The solubility of gas is inversely proportional to the temperature. So if the temperature of the body is increased, solubility of gas decreases.

5. As we know, the solubility of gas is inversely proportional to the temperature. So if the temperature of the body is increased, solubility of gas decreases.

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135

2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: solubility, gas, increase, decrease, temperature, partial pressure, diver, air, change, environment, same, oxygen, anesthetic gas, 10 m, 1 atm.

Andre45 [30]3 years ago
6 0
<span>1. the pressure of a gas over a solvent is increase
</span> Gas solubility decreases"<span>

2. the partial pressure of an anesthetic gas is increased 
</span> " gas solubility does not change"<span>

3. air in blood a diver descends 10 M and pressure increases by 1 atm
</span> Gas solubility decreases"<span>

4. the temp is increase
</span>"gas solubility increases"<span>

5. O2 the temp of a body of water rises.
</span>"gas solubility increases"

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Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.
olasank [31]
Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1 
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)

<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

<span>Then the mass of the solvent is:
</span>
m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{m_1}{M_1*m_2}

\omega =  \dfrac{200}{132*0,821}

\omega =  \dfrac{200}{108,372}

\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

_________________________________
_________________________________


<span>Another way to find the answer:
</span>
We have the following data: 

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4 
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol 


<span>Let's find the number of mols (n), let's see:

</span>n =  \dfrac{m_1}{M_1}

n = \dfrac{200}{132}

n \approx 1,5\:mol

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{n}{m_2}

\omega =  \dfrac{1,5}{0,8}

&#10;\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

I hope this helps. =)
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