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amm1812
3 years ago
5

Question 5

Chemistry
1 answer:
baherus [9]3 years ago
5 0

Answer:

2.03 atm

Explanation:

Number of moles of He = 1g/4g/mol = 0.25 moles

Number of moles of F2 = 14.0g/38 g/mol = 0.37 moles

Number of moles of Ar=19.0g/40g/mol = 0.48 moles

Total number of moles = 0.25 + 0.37 + 0.48 = 1.1 moles

From;

PV=nRT

P= pressure of the gas mixture

V= volume of the gas mixture

n= total number of moles of the gas mixture

R= gas constant

T= temperature of the gas mixture

P= nRT/V

P= 1.1 × 0.082 × 293/13

P= 2.03 atm

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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
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Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

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\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

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Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

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c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

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

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

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