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MaRussiya [10]
3 years ago
5

A balloon of helium was put in the freezer at -23°C. Its volume at this temperature was 2.5 liters. It was removed from the free

zer. Eventually, it reached a temperature of 177°C. What would its volume be at this temperature? (Neglect any force used to stretch the rubber balloon.) (1.)45000 (2.) 0.22 (3.) 4.5 (4.) 1.4
Chemistry
1 answer:
kotegsom [21]3 years ago
3 0

4.5 L

Explanation:

We will use the combined gas law formula;

P₁V₁/T₁ = P₂V₂/T₂

Because the pressure remains the same, it cancels out and what remains is;

V₁/T₁ = V₂/T₂

Before evaluating remember to change the temperatures into Kelvin;

-23° = 250.15 K

177° = 450.15 K

We replace the variables with known values;

2.5 / 250.15 = V₂ / 450.15

V₂ = 2.5 /250.15 * 450.15

V₂ = 4.4988 L

≅ 4.5 L

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Determination of sign of w

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In the given case, work is done on the system, therefore sign of w is positive, or w > 0

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HACTEHA [7]

The molarity of the solution will be 3.45 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂  

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

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345 mL of a 2.5 M NaCl solution.

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V₁=345 ml

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\tt M_2=\dfrac{M_1.V_1}{V_2}\\\\M_2=\dfrac{2.5\times 345}{250}\\\\M_2=3.45~M

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