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Lubov Fominskaja [6]
3 years ago
15

A gas at a pressure of 360 torrs and a volume of 750 ml. If the pressure becomes 1.5 atm what is the new volume?

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

V₂ = 236.84 mL

Explanation:

The relation between pressure and volume is inverse.

We can write it as follows :

P_1V_1=P_2V_2

We have,

P₁ = 360 torrs, V₁ = 750 mL, P₂ = 1.5 atm = 1140 torr.

So,

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{360\times 750}{1140}\\\\V_2=236.84\ mL

So, the new volume of the gas is 236.84 mL.

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The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo
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<u>Answer:</u> The solubility product of silver (I) phosphate is 9.57\times 10^{-10}

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We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L

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The chemical equation for the ionization of silver (I) phosphate follows:

Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)  

                            3s                  s

The expression of K_{sp} for above equation follows:

K_{sp}=(3s)^3\times s

We are given:  

s=2.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}

Hence, the solubility product of silver (I) phosphate is 9.57\times 10^{-10}

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