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I am Lyosha [343]
3 years ago
10

Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh

at was the unknown volume of your 8.61 m h2so4 solution that you began with
Chemistry
1 answer:
olga55 [171]3 years ago
3 0
<span>1.02x10^2 ml Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this m0*v0 = m1*v1 where m0, v0 = molarity and volume of original solution m1, m1 = molarity and volume of final solution. Solve for v0, then substitute the known values and calculate: m0*v0 = m1*v1 v0 = (1.75 M * 500 ml)/8.61 M v0 = (1.75 M * 500 ml)/8.61 M V0 = 101.6260163 Rounding to 3 significant figures gives 102 ml. So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>
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A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap
elena-14-01-66 [18.8K]

Answer:

The empirical formula of the compound = C_4H_8S_1O_1

Explanation:

Mass of carbon dioxide gas = 59.060 mg = 0.059060 g

1 mg = 0.001 g

Moles of carbon dioxide = \frac{0.059060 g}{44 g/mol}=0.0013 mol

Moles of carbon in 0.0013 moles of carbon dioxide gas = 1 × 0.0013 mol = 0.0013 mol

Mass of 0.0013 moles of carbon = 12 g/mol\times 0.0013 mol=0.0156 g

Mass of water = 24.176 mg = 0.024176

Moles of water = \frac{0.024176 g}{18 g/mol}=0.0013 mol

Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol

Mass of 0.0013 moles of hydrogen= 1 g/mol\times 0.0013 mol=0.0013 g

Mass of sulfur dioxide = 20.326 mg = 0.020326 g

Moles of sulfur dioxide = \frac{0.020326 g}{64 g/mol}=0.00032 mol

Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol

Mass of 0.00032 moles of sulfur = 32 g/mol\times 0.00032 mol=0.01024 g

Mass of oxygen in the sample = x

Mass of sample = 33.153 mg = 0.033153 g

0.033153 g = 0.0156  g + 0.0013 g + 0.01024 g + x

x = 0.006013 g

Moles of oxygen = \frac{0.006013 g}{16 g/mol}=0.00038 mol

For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:

Carbon : \frac{0.0013 mol}{0.00032 mol}=4

Hydrogen: \frac{0.0026 mol}{0.00032 mol}=8

Sulfur : \frac{0.00032 mol}{0.00032 mol}=1

Oxygen : \frac{0.00038 mol}{0.00032 mol}=1

The empirical formula of the compound = C_4H_8S_1O_1

8 0
4 years ago
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