Answer:
0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.
Answer:
0.23 V.
Explanation:
<em>∵ ΔG° = -RT lnK.</em>
∴ ΔG° = -RTlnK = -(8.314 J/mol)(298 K) ln(7.3 × 10⁷) = - 44.86 x 10³ J/mol.
<em>∵ ΔG° = - nFE°</em>
∴ E° = - ΔG°/nF = - (- 44.86 x 10³ J/mol)/(2 x 96500 s.A/mol) = 0.2324 V ≅ 0.23 V.
B is the answer
I double checked hope it helps
Answer:
4.07L of a 0.110M NaF are needed
Explanation:
Based on the reaction:
SrCl₂(aq)+2NaF(aq)⟶SrF₂(s)+2NaCl(aq)
<em>1 mole of strontium chloride react with 2 moles of NaF</em>
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361mL of 0.620M SrCl₂ solution has:
0.361L ₓ (0.620mol / L) = 0.22382 moles SrCl₂.
Moles of NaF for a complete reaction must be:
0.22382 moles SrCl₂ ₓ (2 mol NaF / 1 mol SrCl₂) = <em>0.44764 moles of NaF</em>
If you have a solution of 0.110M NaF, the moles of NaF needed are:
0.44764 moles of NaF ₓ (1L / 0.110mol NaF) = <em>4.07L of a 0.110M NaF are needed</em>
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