Answer:
Molar concentration of CO₂ in equilibrium is 0.17996M
Explanation:
Based on the reaction:
NiO(s) + CO(g) ⇆ Ni(s) + CO₂(g)
kc is defined as:
kc = [CO₂] / [CO] = 4.0x10³ <em>(1)</em>
As initial concentration of CO is 0.18M, the concentrations in equilibrium are:
[CO] = 0.18000M - x
[CO₂] = x
Replacing in (1):
4.0x10³ = x / (0.18000-x)
720 - 4000x = x
720 = 4001x
x = 0.17996
Thus, concentrations in equilibrium are:
[CO] = 0.18000M - 0.17996 = 4.0x10⁻⁵
[CO₂] = x = <em>0.17996M</em>
<em></em>
Thus, <em>molar concentration of CO₂ in equilibrium is 0.17996M</em>
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I hope it helps!
Answer:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
b.
c.
d.
Regards!
OK so basically many experiments were done and Ernest Rutherford found that atoms had a positive charge and it contains most of its mass. and for the electron J.J. Thomson found that atoms have a negative charge after a various amount of study and experimentation
Answer:
From the general gas lawa
PV/T=k---constant
Phase 1
735×1.2/385K=2.3
Phase 2
K(constant)×T/P
2.3×290/651
Volume=1.0246dm³
Answer:
concentrated on headlands projecting into the water
Explanation:
