2 H2 + O2 = 2 H2O
1 mole O2 --------------- 2 moles H2O
13.35 moles O2 -------- ( moles H2O)
moles H2O = 13.35 x 2 / 1
<span>= </span>26.7 moles of H2O
<span>hope this helps!</span>
Answer:
Rate of Ar= 1.65 x mol/h
Explanation:
Here we'll use Graham's Law where
R= rate of effusion and M= molar mass
Let's plug in what we know.
You could switch them where Xe is M1 (on the bottom) instead of M2, but I find it easier when the unknown is R1, where it acts like a whole number. It makes the algebra part of the equation easier.
Let's solve for R1.
R1= 1.089 x
R1= 1.65 x mol/h
This question is quite vague, as the initial concentration of ethanol is not provided. However, from experience I can tell you that most laboratory work is done with 98% ethanol, and not absolute ethanol (100%). So in order to calculate the final concentration, we need to take the given values, which includes the initial concentration (98%), the initial volume (50.0mL) and the final volume (100.0mL). We apply the following equation to calculate the final concentration:
C1V1 = C2V2
C1 = Initial concentration
C2 = Final concentration
V1 = Initial volume
V2 = Final volume
(98%)(50.0mL) = (C2)(100.0mL)
Therefore, the final concentration (C2) = 49%