<u>Answer:</u> The value of
for
reaction is 
<u>Explanation:</u>
We are given:
Initial moles of nitrogen gas = 1.30 moles
Initial moles of hydrogen gas = 1.65 moles
Equilibrium moles of ammonia = 0.100 moles
Volume of the container = 1.00 L
For the given chemical equation:

<u>Initial:</u> 1.30 1.65
<u>At eqllm:</u> 1.30-x 1.65-3x 2x
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
Equilibrium moles of nitrogen gas = 
Equilibrium moles of hydrogen gas = 
Putting values in above expression, we get:

Calculating the
for the given chemical equation:


Hence, the value of
for
reaction is 
Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which
is the <u>acid</u> and
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.
Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains
number of particles.
We know that:
Charge on 1 electron = 
Charge on 1 mole of electrons = 

is passed to deposit = 1 mole of copper
63.5 g of copper is deposited by = 193000 C
of copper is deposited by =
To calculate the time required, we use the equation:

where,
I = current passed = 40.0 A
q = total charge = 42551181 C
t = time required = ?
Putting values in above equation, we get:

Converting this into hours, we use the conversion factor:
1 hr = 3600 seconds
So, 
Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours
Answer:
Ok so, b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...
i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced
ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.
iii. Calculate the standard potential (voltage) of the cell
Look up the reduction potential,
E
⁰
red
, for the reduction half-reaction in a table of reduction potentials
Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,
E
⁰
ox
=
-
E
⁰
red
.
iv. What kind of electrochemical cell is this? Explain your answer.
All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells
1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery
Explanation: