0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
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The volume of HCl required is 23.89 mL
Calculation of volume:
The reaction:

As HCl and NaOH react in 1 : 1 ratio.
Volume of NaOH= 718 mL
Concentration= 0.183M
Volume of HCl= ?
Concentration= 5.50M
Using the dilution formula:


Therefore,
Volume of HCl required will be 23.89 mL.
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A electrochemical reaction is said to be spontaneous, if
Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)
The cell representation of above reaction is given by;
</span>

Hence,

we know that,

and

Therefore,

= - 0.25 - (-0.47) = 0.22 v
Since,
is positive, hence cell reaction is spontaneous
.....................................................................................................................
Answer 2: Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;

Hence,

we know that,

and

Therefore,

= - 0.126 - 0 = -0.126 v
Since,
is negative, hence cell reaction is non-spontaneous.....................................................................................................................
Answer 3:
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;

Hence,

we know that,

and

Therefore,

= - 0.22 - (-0.913) = 0.693 v
Since,
is positive, hence cell reaction is spontaneous