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MA_775_DIABLO [31]
3 years ago
12

What is the empirical formula for C 57.66

Chemistry
1 answer:
ELEN [110]3 years ago
4 0

C5H12O2 (2-ISOPROPOXYETHANOL)

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Anastaziya [24]

Answer:

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Explanation:

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The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt
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Molarity is expressed as the number of moles of solute per volume of the solution. For example, we are given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH per 1 L volume of the solution. To calculate the moles of NaCl in 1.0 M of solution, we simply multiply the volume given of the solution.

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7 0
3 years ago
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(Thermodynamics)
frutty [35]

Answer:

3853 g

Step-by-step explanation:

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1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

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