The molar mass of a, b and c at STP is calculated as below
At STP T is always= 273 Kelvin and ,P= 1.0 atm
by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation
that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula
that is molar mass = (mass x RT)/ PV
density is always = mass/volume
therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K
the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol
the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol
the molar mass of c
=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol
therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol
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(don’t count on my answer but I think it might be this one and I apologize if you get it wrong)
Answer:
The answer to your question is: 24 grams of D
Explanation:
To answer this question we need to remember the Lavoisier law of conservation of mass, which says that in a chemical reaction matter is neither created nor destroyed.
This means that the amount of matter stays the same.
Then, the reaction is
A + B ⇒ C + D
26 g 12 g 14 g x
mass
of reactants 38 g ? mass of products, but it must be
equal to the mass of products
Then 14g + x = 38
x = 38 - 14
x = 24 g of D
Answer:
4 boiling point elevation
