<span>This problem uses the
relationship between Ka and the concentrations of the ions. Calculations are as follows:</span>
<span>
</span><span>1.9 x 10-5</span>= x^2 / (0.25 - x)
<span>x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3
</span><span>pH = -log (2.2 x 10^-3)</span> = 2.66
Idk but you can try to look it up. On google or somewhere else but yea idk about that one
1) Balanced chemical equation:
2C2H2 + 5O2 ----> 4CO2 + 2H2O
2) State the molar ratios
2 mole C2H2 : 5 mole O2 : 4 mol CO2 : 2 mole H2O
3) Use the periodic table to get the atomic masses of each element and from that the molar masses of O2 and C2H2
Atomic masses:
O: 16 g / mol
C: 12 g / mol
H: 1 g / mol
Molar masses of O2 and C2H2:
O2: 2 * 16 g/mol = 32 g/mol
C2H2: 2*12 g/mol + 2*1g/mol = 26 g / mol
3) Convert 859.0 grams of C2H2 into moles:
859.9 g / 26.0 g /mol = 33.07 mol C2H2
4) State the molar proportion with C2H2 and O2
x mol O2 / 33.07 mol C2H2 = 5 mol O2 / 2 mol C2H2
=> x = 33.07 mol C2H2 * 5 mol O2 / 2 mol C2H2 = 82.68 mol O2
5) Convert 82.68 mol O2 to grams
82.68 mol * 16 g/mol = 1322.9 g O2.
Answer: 1322.9 grams of O2
HNO3 + Ca —-> H2O + N20 + Ca(NO3)2
