You must use 2.50 mL of the concentrated solution to make 10.0 mL of the dilute solution.
We can use the dilution formula
<em>V</em>_1<em>C</em>_1 = <em>V</em>_2<em>C</em>_2
where
<em>V</em> represents the volumes and
<em>C</em> represents the concentrations
We can rearrange the formula to get
<em>V</em>_2 = <em>V</em>_1 × (<em>C</em>_1/<em>C</em>_2)
<em>V</em>_1 = 10.0 mL; <em>C</em>_1 = 5.00 g/100. mL
<em>V</em>_2 = ?; ____<em>C</em>_2 = 20.0 g/100. mL
∴ <em>V</em>_2 = 10.0 mL × [(5.00 g/100. mL)/(20.0 g/100. mL)] = 10.0 mL × 0.250
= 2.50 mL
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<span>Reaction of ethane combustion:
2C2H6 (g) + 7O2 (g) ----> 4CO2 (g) + 6H2O
According to the reaction, we can see that </span>C2H6 and CO2 have following stoichiometric ratio:
n(C2H6) : n(CO2) = 2 : 4
If we know the number of moles of ethane we can calculate moles of carbon dioxide:
2 x n(CO2) = 4 x n(C2H6)
n(CO2) = 2 x n(C2H6)
n(CO2) = 2 x 5.4 = 10.8 mole of CO2 <span>are produced</span>
Answer:
86.2 or 431/5
Explanation:
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