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Oduvanchick [21]
3 years ago
12

How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?

Chemistry
1 answer:
mars1129 [50]3 years ago
3 0

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

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Mole percent of O2 = 10% = 0.1 moles
Mole percent of N2 = 10% = 0.1 moles
Mole percent of He = 80% = 0.8 moles
Molar Mass of O2 = (2 x 16) x 0.1 = 3.2
Molar Mass of N2 = (2 x 14) x 0.1 = 2.8
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1. Molar Mass of the mixture = 3.2 + 2.8 + 3.2 = 9.2 grams
2. Since at constant volume density is proportional to mass, so the ratio of
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What material is least likely to be recognized as a mixture by looking under a microscope
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What would the expected temperature change be (in Fahrenheit) ida 0.5 gran sample of water released 50.1 J of heat energy? The s
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<h3>Answer:</h3>

23.95 °C

<h3>Explanation:</h3>

We are given;

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  • Specific heat capacity is 4.184 J/g°C

We are required to calculate the change in temperature;

  • Quantity of heat absorbed is given by the formula;
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That is, Q = mcΔT

Rearranging the formula;

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Therefore;

ΔT = 50.1 J ÷ (0.5 g × 4.184 J/g°C)

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