Answer:
pOH=9.9
Explanation:
pH=-log[H+]= -log[0.0000877]
=4.06
pOH+ pH=14
pOH=14-4.06= 9.91
The molecular formula of methylpropan-1-ol is C4H10O, so the complete combustion equation is: C4H10O + 6O2 --> 4CO2 + 5H2O. This mean to completely combust 1.0mol of methylpropan-1-ol, 6 mol of O2 is required. Molar mass of O2 is 32 g/mol, so 32g/mol x 6mol = 192 g of O2 is required. At room temperature and pressure, the density of O2 is 1.3315 g/L (this can be obtained by density of gas = P/RT). So the volume of O2 = mass/density = 192g/1.3315(g/L) = 144 L = 144 dm3. The answer is B.
Answer:
ultraviolet .rays....hope am correct
