The mass number of aluminium hydroxide is 78 thus, the number of moles in 0.745 g is:
no. of moles= mass/ RFM
= 0.745/78
=0.00955moles
Therefore the 0.00955 moles should be in the 35.18 ml
therefore 1000ml of the solution will have:
(0.00955ml×1000ml)/35.18
=0.2715moles
The solution will be 0.27M hydrochloric acid
2H₂ + O₂ = 2H₂O
n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol
n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol
H₂ : O₂ = 2 : 1
2.5 : 1.25 = 2 : 1
n(H₂O)=n(H₂)=2n(O₂)=2.5 mol
m(H₂O)=n(H₂O)M(H₂O)
m(H₂O)=2.5mol*18.0g/mol=45.0 g
Answer:
The answers are A,B,C.
Explanation: Just got it right on Edge 2020
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.
![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)
Regards!
Answer:
1, 2, 1
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!