Answer:
Explanation:
according to balance chemical equation
3 A2 moles produced 2 moles of A3B
so 12 moles A2 will produced moles of A3B= 12*2/3=24/3= 8
therefore 12 moles of A2 produced 8 moles of A3B
Answer:
b) 49.48% C, 5.19% H, 28.85% N, and 16.48% O
Explanation:
we find the mass for each element in one mole by multiplying the number of atoms in one molecule with the atomic mass
mC=8Ac=8*12=96g
mH=10AH=10*1=10g
mN=4AN=4*14=56g
mO=2AO=2*16=32g
by adding the masses together we find the molar mass of the molecule
M=mC+nH+mN+mO=96+10+56+32=194g/mole
we apply the rule of threes to find the percentage of each element
194g..96gC..10gH...56gN....32gO
100g....a...........b...........c.............d
a=(100*96)/194=49.48%C
b=(100*10)/194=5.19%H
c=(100*56)/194=28.85%N
d=(100*32)/194=16.48%O
Answer:
15.6g Ag2SO4
Explanation:
2AgNO3 + H2SO4 --> Ag2SO4 + 2HNO3
-2x -x
0.1-2x. 0.155-x
x=0.05 x=0.155
0.05mol Ag2SO4 x 311.78g = 15.6g Ag2SO4
Answer:
The 
for the reaction 
will be 4.69.
Explanation:
The given equation is A(B) = 2B(g)
to evaluate equilibrium constant for
![K_c=[B]^2[A]](https://tex.z-dn.net/?f=K_c%3D%5BB%5D%5E2%5BA%5D)
= 0.045
The reverse will be 
Then, ![K_c = \frac{[A]}{[B]^2}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BA%5D%7D%7B%5BB%5D%5E2%7D)
= 
= 
The equilibrium constant for will be
= 4.69
Therefore, for the reaction will be 4.69.
