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
Answer:d
Explanation:I’m not 100% sure how to explain it but I’m almost for sure it’s d.
Answer:
Explanation:
To find the weight (W) of the pond contents first we need to use the following equation:
(1)
Where m the mass and g is the gravity
Also, we have that the mass is:
(2)
Where ρ is the density and V the volume
We cand calculate the volume as follows:
(3)
Where L is the length, w is the wide and d is the depth
By entering equation (2) and (3) into (1) we have:
Therefore, the weight of the pond is 6.65x10⁶ lbf.
I hope it helps you!
Explanation:
put them in where the letters start with
