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Ainat [17]
3 years ago
10

Give the half-equation for the reaction at the anode, during

Chemistry
1 answer:
oee [108]3 years ago
3 0
The answer would be B
You might be interested in
1. Complete the balanced dissociation equation for the compound below. If the compound does not dissociate, write NR after the r
puteri [66]

Answer:

1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Explanation:

The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:

(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).

The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.

S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)

Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.

2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)

The net ionic equation is given below:

Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

7 0
3 years ago
Which identifies an oxidation-reduction reaction?
Strike441 [17]

Answer:

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄  →  Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is  also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized

8 0
3 years ago
We dissolve 93 grams of sodium sulfate (Na2SO4) in 20 grams of water. What is the percent by mass of sodium sulfate in this solu
dlinn [17]
It should be for the total solution of 93 plus 20 grams which is 113 grams so 93 divided by 113 grams comes to 82.3% sodium sulfate and this can be checked by multiplying 113 grams by 0.823 which results in 93 grams of sodium sulphate. 
4 0
4 years ago
What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g
Anika [276]
To get the percent yield, we will use this formula: ((Actual Yield)/(Theoretical Yield)) * 100% Values given: actual yield is 220.0 g theoretical yield is 275.6 g Now, let us substitute the values given. (220.0 grams)/(275.6 grams) = 0.7983 Then, to get the percentage, multiply the quotient by 100. 0.7983 (100) = 79.83% Among the choices, the most plausible answer is 79.8% <span>
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4 0
4 years ago
Read 2 more answers
When sodium is put in water, the metal floats on the surface and reacts to form sodium
mrs_skeptik [129]

Answer: Sodium also floats on the surface, but enough heat is given off to melt the sodium (sodium has a lower melting point than lithium and the reaction produces heat faster) and it melts almost at once to form a small silvery ball that dashes around the surface.

Explanation:

8 0
3 years ago
Read 2 more answers
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