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MaRussiya [10]
4 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]4 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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Answer:

Approximately 19.1\; \rm g.

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<h3>Number of moles of formula units of magnesium sulfate required to make the solution</h3>

The unit of concentration in this question is "\rm M". That's equivalent to "\rm mol \cdot L^{-1}" (moles per liter.) In other words:

c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}.

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L.

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\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}.

<h3>Mass of magnesium sulfate in the solution</h3>

Look up the relative atomic mass data of \rm Mg, \rm S, and \rm O on a modern periodic table:

  • \rm Mg: 24.305.
  • \rm S: \rm 32.06.
  • \rm O: 15.999.

Calculate the formula mass of \rm MgSO_4 using these values:

M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}.

Using this formula mass, calculate the mass of that (approximately) 0.159\; \rm mol of \rm MgSO_4 formula units:

\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}.

Therefore, the mass of \rm MgSO_4 required to make this solution would be approximately 19.1\; \rm g.

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