<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
and 
where, x = mole fraction in solution
= pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.


,
,





Thus (1-x0 = (1-0.34)=0.66
Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66
The answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---> Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
Hf C2H2 = 226.77
Hf Ca(OH)2 = -986.2
<span>Hf H2O = -285.83
Now,
</span><span>add them up. 226.77 - 986.2 + (2*285.83) = -187.77
</span><span>Add back the total enthalpy that is given in the question
-187.77+127.2 = -60.57 </span>