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Rufina [12.5K]
2 years ago
8

Consider the half reactions below for a chemical reaction.

Chemistry
2 answers:
frozen [14]2 years ago
6 0

Answer:

IT's d on edge

Explanation:

Butoxors [25]2 years ago
4 0

Upper M g (s) plus 2 upper H right arrow upper M g (a q) plus upper H subscript 2 (g).

 Mg + 2H⁺      →      Mg²⁺ +    H₂

Explanation:

The net ionic equation is the sum of the oxidation half reactions.

The half-reactions are:

 Mg   →   Mg²⁺    +   2e⁻   Oxidation half reaction

  2H⁺    +     2e⁻    →       H₂    reduction half reaction

Combining the reactions gives:

Mg + 2H⁺    +     2e⁻   →       H₂ + Mg²⁺    +   2e⁻  

Since the electrons are balanced, they cancel out of the equation,

 This leaves behind;

   Mg + 2H⁺      →      Mg²⁺ +    H₂

   

Solution:

Ionic equation brainly.com/question/2947744

#learnwithBrainly

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15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno
timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

n=0.050 M\times 0.0150L=0.00075 mol

Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)

Moles of barium ions: 1\times 0.00075 mol=0.00075 mol

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

n'=0.10 M\times 0.1000 L=0.01 mol

KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)

Moles of iodate ions = 1\times 0.01 mol=0.01 mol

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M

Concentration of iodate ions:

[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M

Solubility product of barium iodate,K_{sp}=4.01\times 10^{-9}

Ionic product of the barium iodate in solution :K_i

Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}

K_i=[Ba^{2+}][IO_2^{-}]^2

K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}

K_{sp}  ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

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