Answer:
Explanation:
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In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:
Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:
Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:
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Answer:
Ne
Explanation:
Noble gas, Unreactive therefore does not exist as a diatomic molecule
Answer:
a) 7.0.
b) Nickel sulfate hepta hydrate.
c) 280.83 g/mol.
d) 44.9%.
Explanation:
<u><em>a) What is the formula of the hydrate?</em></u>
The mass of the hydrated sample (NiSO₄.xH₂O) = 5.0 g,
The mass of the anhydrous salt (NiSO₄) = 2.755 g,
The mass of water = 5.0 g - 2.755 g = 2.245 g.
∴ no. of moles of water = mass/molar mass = (2.245 g)/(18.0 g/mol) = 0.1247 mol.
∴ no. of moles of anhydrous salt (NiSO₄) = mass/molar mass = (2.755 g)/(154.75 g/mol) = 0.0178 mol.
∴ water of crystallization in the sample (x) = no. of moles of water/no. of moles of anhydrous salt (NiSO₄) = (0.1247 mol)/(0.0178 mol) = 7.0.
<u><em>b) What is the full chemical name for the hydrate?</em></u>
The name of the salt (NiSO₄.7H₂O) is Nickel sulfate hepta hydrate.
<u><em>c) What is the molar mass of the hydrate? </em></u>
(NiSO₄.7H₂O)
The molar mass = molar mass of NiSO₄ + 7(molar mass of H₂O) = (154.75 g/mol) + 7(18.0 g/mol) = 280.83 g/mol.
<em><u>d) What is the mass % of water in the hydrate?</u></em>
The mass % of water = (mass of water)/(mass of hydrated sample) x 100 = (2.245 g)/(5.0 g) x 100 = 44.9%.
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
