4. If you evaporate the water first, then the sand and salt will be nearly impossible to separate. That eliminates 1 and 2. You can't filter salt out from an aqueous solution, so that eliminates 3. 4 works because you can filter out the sand (which doesn't dissolve) then evaporate the water away from the salt.
Answer:

Explanation:
Hello,
In this case, given the amounts of water and carbon dioxide we should invert the given reaction as hydrogen will be producted rather than consumed:

Consequently, the equilibrium constant is also inverted:

In such a way, we can now propose the law of mass action:
![Kc'=\frac{[H_2][CO_2]}{[H_2O][CO]}](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B%5BH_2%5D%5BCO_2%5D%7D%7B%5BH_2O%5D%5BCO%5D%7D)
And we can express it in terms of the initial concentrations of the reactants and the change
due to the reaction extent:
![Kc'=\frac{(x)(x)}{([H_2O]_0-x)([CO]_0-x)}=1.87](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B%28x%29%28x%29%7D%7B%28%5BH_2O%5D_0-x%29%28%5BCO%5D_0-x%29%7D%3D1.87)
Thus, we compute the initial concentration which are same, since equal amount of moles are given:
![[H_2O]_0=[CO]_0=\frac{0.680mol}{70.0L}=0.0097M](https://tex.z-dn.net/?f=%5BH_2O%5D_0%3D%5BCO%5D_0%3D%5Cfrac%7B0.680mol%7D%7B70.0L%7D%3D0.0097M)
Hence, solving for
by using the quardratic equation or solver, we obtain:

For which the correct value is 0.00561M since the other one will produce negative concentrations of water and carbon monoxide at equilibrium. Therefore, the number of moles of hydrogen at equilibrium for the same 70.0-L container turn out:

Best regards.
Answer:
acid is like sour taste
and bases increases the H for concentration of gas.
Answer:
Carbohydrates= energy that is made up of glucose
Explanation:
Protein= last one
liqid = insulate you