A) Earlier you were told that of all of the 14 solutes you will be studying, the only one that is not appreciably ionized in wat
1 answer:
Answer:
Throughout the overview section following portion, the description and according to particular circumstance is defined.
Explanation:
As per the question,
⇒
- A weak basis seems to be NH3. It serves as a base since the aqueous solution or phase is protonated. But NH3 +, just becoming a weak base, is therefore deprotonated into form NH3, and therefore also 90% of ammonia becomes found throughout NH3 state in aqueous solution.
⇒
However, it is also available in NH3 form throughout the aqueous solution much of the moment.
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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
The equilibrium constant for the reaction is 0.00662
Explanation:
The balanced chemical equation is :
2NO2(g)⇌2NO(g)+O2(g
At t=t 1-2x ⇔ 2x + x moles
The ideal gas law equation will be used here
PV=nRT
here n= =
= density
P = density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm
putting the values in reaction
0.75 =
M = 34.61
to calculate the Kc
Kc=
x M NO2 +
M NO+
M O2
Putting the values as molecular weight of NO2, NO,O2
34.61=
x= 0.33
Kc=
putting the values in the above equation
Kc = 0.00662