Let's look at the molar weight of the answers:
NO is 30 g/mol
NO2 is 46
N2O is 44
N2O4 is 124
<span>We have the grams of the product, so we need the moles in order to calculate the molar weight. We us PV=nRT for this, assuming standard temperature and pressure. </span>
You were given the liters (.120L)
Std pressure is 1 atmosphere
You're looking for n, the number of moles
<span>Temp is 293.15 kelvin, thats standard </span>
And r is the gas constant in liters-atm per mol kelvin
(.120 liters)(1atm)=n(293.15K)(.08206)
Solving for n is .0049883835 mol
<span>.23g divided by .0049883 mol is about 46g/mol. You're answer is B I think, NO2
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Answer:2.86x10^-7m
Explanation:E=hc/^
E=6.94x10^-19J
c = 2.9979x10^8m/s
h= 6.626x10^-34Js
^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19
= 2.86x10^-7m
Convert 57.6 L to dm3 and divide it by 24
We know that the molar mass of N is 14 and O is 16,
therefore the molar mass of N2O is:
molar mass N2O = 14 * 2 + 16 = 44 g/mol
The number of moles:
moles N2O = 0.187 / 44
moles N2O = 0.00425 mol
There are 2 moles of N per 1 mole of N2O hence:
moles N = 0.00425mol * 2
<span>moles N = 0.0085 mol</span>
Since both samples are pure CH4 (methane), the proportion of C to H that evolves from the decomposition should be equal. In equation form:
35.0 g C / 2.04 g H = 23.0 g C / x g H
Solving for x gives a value of x = 1.3406 g H
So 1.3406 grams of hydrogen will be produced from sample b.
