Answer:
3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)
Explanation:
When aqueous solutions of chromium(II) iodide and sodium phosphate are combined, solid chromium(II) phosphate and a solution of sodium iodide are formed. The molecular equation is:
3 CrI₂(aq) + 2 Na₃PO₄(aq) ⇄ Cr₃(PO₄)₂(s) + 6 NaI(aq)
The full ionic equation includes all the ions and the molecular species.
3 Cr²⁺(aq) + 6 I⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 I⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)
Answer:
The limiting reagent is the SF₄
Explanation:
In order to determine the limiting reagent we convert the mass of the reactants to moles, and then, we work with stoichiometry.
4.687 g / 108.06 g/mol = 0.0433 moles of SF₄
6.281 g / 333.8 g/mol = 0.0188 moles of I₂O₅
The reaction is: 5SF₄ + 2I₂O₅ = 4IF₅ + 5SO₂
Ratio is 5:2. 5 moles of fluoride react with 2 moles of pentoxide
Then, 0.0433 moles of fluoride will react with (0.0433 . 2) / 5 = 0.0173 moles
We have 0.0188 moles of I₂O₅ and we need 0.0173 so there are some moles of pentoxide that remains after the reaction. In conclussion, the limiting reagent is the SF₄. We verify:
2 moles of pentoxide react with 5 moles of SF₄
Therefore, 0.0188 moles of I₂O₅ will react with (0.0188 . 5) / 2 = 0.0470 moles.
As we have 0.0433 moles of SF₄, we do not have enough moles.
Answer is: <span> the equilibrium concentration of Br</span>₂ is 0,02 mol/L.<span>
</span>Chemical reaction: Br₂ + Cl₂ → 2BrCl.
Kc = 7,0.
c₀(Br₂) = 0,25 mol ÷ 3 L.
c₀(Br₂) = 0,083 mol/L.
c₀(Cl₂) = 0,55 mol ÷ 3 L.
c₀(Cl₂) = 0,183 mol/L.
Kc = c(BrCl)² ÷ c(Br₂) · c(Cl₂).
7 = (2x)² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
7 = 4x² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
Solve q<span>uadratic equation: x = 0,063 mol/L.
</span>c(Br₂) = 0,083 mol/L - 0,063 mol/L = 0,02 mol/L.
Answer:
Compound 3 is a clear liquid with a strong pleasantly fruity smell. If cooled it freezes at about −10°C. In the solid state it does not conduct electricity. ... It dissolves slightly in water, and a solution of 2g in 100mL of water doesn't change the electrical conductivity of the water.