Explanation:
The substance that is in excess that doesn't get used up as a reactant is called limiting reactant.
The answer is 236.5 J/K
According to Δ G formula:
ΔG = ΔH - TΔS
when ΔG is the change in free energy (KJ)
and ΔH is the change in enthalpy (KJ)= ΔHvap * moles
= 71.8 KJ/mol * 1.11 mol
= 79.7 KJ
and T is the absolute temperature (K)= 64 °C + 273°C = 337 K
Δ S is the change in entropy KJ/K
by substitution:
when at equilibrium ΔG = 0
∴ΔS = ΔH / T
=79.7 KJ/ 337 K
= 0.2365 KJ/K
= 236.5 J/K
Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
<em></em>
Answer:
<u>Explanation</u>:
<u>Number of molecules for
</u>

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol



<u>Number of molecules for for
</u>

= Atomic mass of 3(Na) + P + 4(O)
= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

