Answer:

Explanation:
Hello!
In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

Thus, the percent water is:

So we plug in to obtain:

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Answer:- 
Explanations:- The solution we have is a buffer solution and we know that a buffer solution resists a change in its pH if a strong acid or base is added to it.
Here, the buffer solution we have is of a weak base and it's conjugate acid. So, a strong acid(nitric acid) is added to this buffer then it reacts with the base present in the buffer so that the acid could be neutralized. This is called buffer action.
The net ionic equation is written as:

Note that
is a strong acid and nitrate ion is the spectator ion so it is not included in the net ionic equation.
2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
Is a solid because liquids are not is not that close and gas is very far apart.