The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
Answer:
5.63 mol.
Explanation:
- The balanced chemical equation between NO₂ and H₂O is:
<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g),
</em>
It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.
<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>
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- To find the no. of moles of HNO₃ produced:
3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.
8.44 mol of NO₂ produces → ??? mol of HNO₃.
∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.
Answer:
1.30mL
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH —> K2SO4 + 2H2O
From the equation above we obtained the following:
Mole of acid (nA) = 1
Mole of base (nB) = 2
The following data were obtained from the question:
Mb = 0.00945M
Vb =?
Va = 50mL
Ma = 1.23 × 10^−4M
Using MaVa / MbVb = nA/nB, we can calculate the volume of KOH as illustrated below:
MaVa / MbVb = nA/nB
(1.23 × 10^−4 x 50)/0.00945xVb = 1/2
Cross multiply to express in linear form
1.23 × 10^−4 x 50 x 2 = 0.00945xVb
Divide both side by 0.00945
Vb = (1.23 × 10^−4 x 50 x 2) /0.00945
Vb = 1.30mL
