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3 years ago

the pressure on 2.50L of N20 changes from 105 kPa to 40.5 kPa. if the temperature does not change, what will the new volume be.

Chemistry

1 answer:

tresset_1 [31]3 years ago

6 0

Answer:

6.48L

Explanation:

Given parameters:

V₁ = 2.5L

P₁ = 105 kPa

P₂ = 40.5 kPa

Condition: constant temperature

Unknown:

V₂ = ?

Solution:

To solve this problem, we are considering pressure and volume relationship. This should be solved by applying the knowledge of Boyle's law.

The law states that "The volume of fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

Mathematically;

P₁V₁ = P₂V₂

where P and V are pressure and volume, 1 and 2 represents initial and final states.

Substitute to find the V₂;

105 x 2.5 = 40.5 x V₂

Solving for V₂ gives 6.48L

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Answer:

The compound amount is $303.03 .

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3 years ago

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Read 2 more answers

171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in o

Fantom [35]

The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?

a. 16mm Hg

b. 24mm Hg

c. 20mm Hg

d. 12mm Hg

Answer: The vapor pressure (VP) of this solution at $25^{o}C$ is closest to the value 24 mm Hg.

Explanation:

Given: Mass of sucrose = 171 g

Mass of water = 1 L = 1000 g

Vapor pressure of water = 24 mm Hg

As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol$

Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol$

Total moles = 55.49 + 0.5 mol = 55.99 mol

Mole fraction of water is as follows.

$Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99$

Formula used to calculate vapor pressure of the solution is as follows.

$P_{i} = P^{o}_{i} \times \chi_{i}$

where,

$P_{i}$ = vapor pressure of component i over the solution

$P^{o}_{i}$ = vapor pressure of pure component i

$\chi_{i}$ = mole fraction of i

Substitute the values into above formula to calculate vapor pressure of water as follows.

$P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\$

Thus, we can conclude that the vapor pressure (VP) of this solution at $25^{o}C$ is closest to the value 24 mm Hg.

4 0

3 years ago

The density of concentrated ammonia, which is 28.0% w/w nh3, is 0.899 g/ml. what volume of this reagent should be diluted to 1.0

vlada-n [284]

Answer: 2.4 ml

Solution :

Molar mass of $NH_3$ = 17 g/mole

Given,: 28% w/w of $NH_3$ solution means 28 g of ammonia in 100 g of solution.

Mass of solution = 100 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{100g}{0.899g/ml}=111.2ml$

Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute $NH_3=\frac{\text {given mass}}{\text {molar mass}}=\frac{28}{17}=1.65moles$

$V_s$ = volume of solution in liter = 0.11 L

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.65moles}{0.11L}=15mole/L$

Using molarity equation:

$M_1V_1=M_2V_2$

$15\times V_1=0.036\times 1.0\times 10^{3}$

$V_1=2.4ml$

6 0

3 years ago