Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺ | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺ + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺ + 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺ + 2e⁻⇄ Cu²⁺ + 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺ ⇄ Cu²⁺ + 2Ag
Answer:
Element 2
Explanation:
If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.
The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.
Answer:
12.33 cal/sec
Explanation:
As we know,
1 Kcal = 1000 cal
So,
0.74 Kcal = X cal
Solving for X,
X = (0.74 Kcal × 1000 cal) ÷ 1 Kcal
X = 740 cal
Also we know that,
1 Minute = 60 Seconds
Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,
= 740 cal / 60 sec
= 12.33 cal/sec
