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%)
Answer:
See explanation
Explanation:
Now , we have the equation of the reaction as;
2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)
This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4
So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.
Answer:
6.24 x 10-3 M
Explanation:
Hello,
In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:
![Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change
, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

Thus, we obtain a quadratic equation whose solution is:

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.
Best regards.
As number of gaseous moles in reactant and prodict are same that is 4
So No change will occur
Answer:
Explanation:
In one of the process, energy is built up from scratch, in the other one, energy is liberated for use by an organism or body.
The first process deals with a metabolic reaction in which energy is liberated:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy
In the above process, energy is liberated when glucose combines with oxygen. The waste products are carbon dioxide and water. This process liberates heat energy which can be used to do work.
In the reverse process:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
This process stores energy in carbon chains as chemical energy. It is this energy that is released in the first process.
Therefore, we can see that the first process liberates energy and the reverse process stores energy.