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Sedaia [141]
3 years ago
14

A measured sample of argon gas has a volume of 20.0 L at a pressure of 660 mm Hg. What is the final volume

Chemistry
1 answer:
valkas [14]3 years ago
3 0

Answer:

C . 24 L

Explanation:

Given data:

Initial volume of gas = 20.0 L

Initial pressure of gas = 660 mmHg

Final volume = ?

Final pressure = 550 mmHg

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

660 mmHg × 20.0 L = 550 mmHg × V₂

V₂ = 13200 mmHg. L/ 550 mmHg

V₂ = 24 L

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Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na+(aq) + SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of s
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<u>Answer:</u> The number of moles of Na_2SO_4 is 0.05 moles.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

We are given:

Molarity of solution = 0.10 mol/L

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0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol

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By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

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8 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 87ml+87ml=174ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

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

m = mass of water = 174 g

T_{final} = final temperature of water = 317.4 K

T_{initial} = initial temperature of metal = 298 K

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

q=174g\times 4.18J/g^oC\times (317.4-298)K

q=14110.008J=14.11KJ

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole

Therefore, the enthalpy of neutralization is, -101.37 KJ

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