The answer & explanation for this question is given in the attachment below.
Answer:
The empirical formula of compound is C₂H₆O.
Explanation:
Given data:
Mass of carbon = 12 g
Mass of hydrogen = 3 g
Mass of oxygen = 8 g
Empirical formula of compound = ?
Solution:
First of all we will calculate the gram atom of each elements.
no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms
no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms
no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms
Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.
C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5
C:H:O = 2 : 6 : 1
The empirical formula of compound will be C₂H₆O
Answer:
Well this is a metathesis or partner exchange reaction....and barium sulfate is as soluble as a brick...
Explanation:
And so...
Ba(NO3)2(aq)+K2SO4(aq)→2KNO3(aq)+BaSO4(s)⏐↓
Note that you simply HAVE TO KNOW that barium sulfate is insoluble....as is lead sulfate, and as is (less so) calcium sulfate
Explanation:
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
