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ycow [4]
3 years ago
9

Consider the redox reaction below.

Chemistry
1 answer:
vovangra [49]3 years ago
4 0

Answer:

Zn(s) → Zn⁺²(aq) + 2e⁻

Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

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6.9 moles of oxygen gas in a fixed volume of 1.33 liters. One of the valves opened and there remains only 1.24 moles oxygen gas.
Romashka [77]

Answer:

The new volume of the gas remains the same. That is new volume of gas is 1.33 litres

Explanation:

This is because gases do not have a definite shape. They therefore take the shape of their containing vessels and hence their volumes are determined by the volume of the container.

For the question above even if some of the gas escapes, as long as there is gas present in the container, its volume remains the same, that is occupies the same space in the container

7 0
2 years ago
Help me
Valentin [98]

Answer:

South American

Explanation:

When you look at a map of plates, only South American forms a boundary with the African plate out of those specific plates

6 0
3 years ago
Read 2 more answers
A mechanic needs a radiator to have 50% antifreeze solution. the radiator currently is filled with 4 gallons of 15% antifreeze s
Zanzabum

<span>let x=gallons of current mixture to be drained and replaced with pure antifreeze.
4-x=gallons of current mixture remaining in the car.</span>

<span>
0.15(4-x)+1.00x=0.50 x 4
0.6-.15x+x=2
0.85x=1.4
x=1.4/0.85 =1.65 gal

Thus, 1.65 gallons of current mixture to be drained and replaced with pure antifreeze.</span>

5 0
3 years ago
At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the
Crank

The decomposition time : 7.69 min ≈ 7.7 min

<h3>Further explanation</h3>

Given

rate constant : 0.029/min

a concentration of  0.050 mol L  to a concentration of 0.040 mol L

Required

the decomposition time

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time

For first-order reaction :

[A]=[Ao]e^(-kt)

or

ln[A]=-kt+ln(A0)

Input the value :

ln(0.040)=-(0.029)t+ln(0.050)

-3.219 = -0.029t -2.996

-0.223 =-0.029t

t=7.69 minutes

4 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
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