Hey there!:
Copper plating on Zinc will occur via this simple reaction:
Cu²⁺ + 2 e⁻ ⇌ Cu
When a species gains electron, it is reduced. Its oxidation state is decreased. For example, in the above reaction Cu2+ gained 2 electron to get reduced to Cu. Its oxidation state changed from +2 to 0. Hence it is a reduction reaction.
So, the correct answer is :
The copper II ions gain two electrons and are reduced.
Answer B
Hope this helps!
A positive acceleration indicates that the object sped up. This means that if you compare the first speed to the second, the second speed should be higher.
A negative acceleration indicates that the object has slowed down. This means that if you compare the first speed to the second, the second speed should be lower.
If an acceleration is 0, it means that it neither slowed down nor sped up.
Now let us analyze your problem by listing down the speed and the time:
At noon: 4 mi/hr
12:30 : 6 mi/hr
2:30 : 2 mi/hr
From noon to 12:30, you will notice that there is an increase in speed. This means that Tommy had a positive acceleration. (Rules out D.)
From 12:30 to 2:30, there is a decrease in speed. This would indicate that Tommy had a negative acceleration. (Rules out C.)
No speed was the same, so acceleration was never 0. (Rules out A.)
From the assumptions above, we can now deduce that the answer is B.
Explanation:
The reactions which are not truly of first order but become reactions of first order under certain conditions are called pseudo first order reactions.
For example, hydrolysis of ester
Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O
Answer:
2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)
Explanation:
Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.
E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V
The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.
