According to Henry's law, solubility of solution is directly proportional to partial pressure thus,
Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.
Putting the values in equation:
On rearranging,
Therefore, solubility will be 1.88 mg of gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.
The empirical formula : MnO₂.
<h3>Further explanation</h3>Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
Mn : O =
The empirical formula : MnO₂
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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