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harina [27]
3 years ago
11

You fill a balloon with helium gas to a volume of 2.68L at 23°C and 789mmHg. Then you release the balloon and it rises. What wil

l the volume be when the pressure drops to 632mm Hg if the temperature remains the same?
Chemistry
1 answer:
wlad13 [49]3 years ago
3 0

Answer : The final volume of balloon will be, 3.35 L

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 789 mmHg

P_2 = final pressure = 632 mmHg

V_1 = initial volume = 2.68 L

V_2 = final volume = ?

Now put all the given values in the above equation, we get:

789mmHg\times 2.68L=632mmHg\times V_2

V_2=3.35L

Therefore, the final volume of balloon will be, 3.35 L

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A solution of 20.0 g of which hydrated salt dissolved in
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(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

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<em>ΔTf = i.Kf.m,</em>

Where, ΔTf is the depression in freezing point of water.

i is van't Hoff factor.

Kf is the molal depression constant.

m is the molality of the solute.

  • Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
  • van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

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MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

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Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

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