The reaction formula CH4 + 2O2 → CO2 + 2H2O shows the oxidation of 1 mole of CH4 (Methane) will yield 1 mole of CO2 (Carbon Dioxide). Since 1 mole of CH4 will weigh 12g (for the Carbon) + 4g (1g for each Hydrogen) = 16g, then 32g of CH4 will correspond to 32g / 16g/mole = 2 moles. Therefore the oxidation of 2 moles of CH4 will yield 2 moles of CO2.
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m
Answer:
for one mole of C2H6 there are 7/2 mole of O2 required. so for4. 50 moles you require 4.50 x 7/2 = 15.75 moles of O2.
Explanation:
i hope it's helpful
Answer:
3Mg(NO3)2(aq)+2Na3PO4(aq)⇒Mg3(PO4)2(s)+6NaNO3(aq)
Explanation:
Answer : The equilibrium constant for this reaction is, 
Explanation :
The given main chemical reaction is:
; 
The intermediate reactions are:
(1) 
; 
(2) 
; 
We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:
(1) 
; 
(2) 
; 
Thus, the equilibrium constant for this reaction will be:
Thus, the equilibrium constant for this reaction is,
