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Artist 52 [7]
2 years ago
5

How many grams of Ar are there in 2.25 moles of Ar?

Chemistry
1 answer:
nydimaria [60]2 years ago
8 0

Answer:

89.88 g

Explanation:

Atomic Mass of Ar: 39.948

Mass = moles * AM

Replacing moles = 2.25 and AM = 39.948 you get the mass of Ar:

Mass = 2.25 * 39.948

Mass = 89.88 g

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What are the symbols of the two elements which are liquid
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Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce
baherus [9]

Answer:

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

<em>Hess's law: ΔH products - ΔH reactants</em>

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<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>

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ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

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7 0
2 years ago
If the solubility of sodium chloride is 36 grams per 100 grams of water, which of the following solutions would be considered un
Murrr4er [49]

Answer: If the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

Explanation:

A solution which contains the maximum amount of solute is called a saturated solution. Whereas a solution in which more amount of solute is able to dissolve is called an unsaturated solution.

Now, the number of moles present in 36 g of NaCl (molar mass = 58.4 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{36 g}{58.4 g/mol}\\= 0.616 mol

This shows that solubility of sodium chloride is 36 grams per 100 grams of water means a maximum of 0.616 mol of NaCl will dissolve in 100 mL of water.

So, a solution in which number of moles of NaCl are less than 0.616 mol per 100 mL then the solution formed will be an unsaturated solution.

  • As 5.8 moles of NaCl dissolved in 1 L (or 1000 mL) of water. So, moles present in 100 mL are calculated as follows.

Moles = \frac{5.8 mol}{1000 mL} \times 100 mL\\= 0.58 mol

  • Moles present in 100 mL of water for 3.25 moles of NaCl dissolved in 500 ml in water are as follows.

Moles = \frac{3.25 mol}{500 mL} \times 100 mL\\0.65 mol

  • Moles present in 100 mL of water for 1.85 moles of NaCl dissolved in 300 ml of water are as follows.

Moles = \frac{1.85 mol}{300 mL} \times 100 mL\\= 0.616 mol

Thus, we can conclude that if the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

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3 years ago
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Answer:

Explanation:

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