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Rufina [12.5K]
4 years ago
11

A solution was prepared by mixing 50.0 g

Chemistry
1 answer:
frez [133]4 years ago
4 0

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

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Answer:

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Explanation:

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Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

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Isolating V in the Ideal Gas Law:

PV=nRT

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V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

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