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asambeis [7]
3 years ago
5

If a chemist has 17.8 moles of N2H4O3, what is the mass of the sample? show work plz

Chemistry
1 answer:
harkovskaia [24]3 years ago
4 0

(\frac{17.8mole}{1} )( \frac{66grams}{1mole} ) \:  \:  \:  \:  17.8 \times 66 = 1.17 \times  {10}^{3} grams
the way that you get 66 is by adding he atomic mass unit of each atom in the formula. so N2 is 14 + 14 H4 is 1 * 4 and 03 is 16 * 3
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Explanation:

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If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
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<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

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<u>Step 3: </u>Calculate moles of SO4^2-

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Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]  

[Ca2+]= 0.050 M   [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

<u> Step 7:</u> Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

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