Answer:
pH = 10.38
Explanation:
∴ molar mass C9H13N = 135.21 g/mol
∴ pKb = - log Kb = 4.2
⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]
∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M
mass balance:
⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)
charge balance:
⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water
⇒ [C9H20O3N+] = [OH-].......(2)
(2) in (1):
⇒ [C9H13N] = 1.516 E-3 - [OH-]
replacing in Kb:
⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])
⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0
⇒ [OH-] = 2.3985 E-4 M
∴ pOH = - Log [OH-]
⇒ pOH = 3.62
⇒ pH = 14 - pOH = 14 - 3.62 = 10.38
In this redox reaction, the Cu goes from oxidation state of (0) to (+2), therefore it oxidises. N in HNO₃ goes from oxidation state of (+5) to N in NO with oxidation state of (+2) and becomes reduced.
Cu acts as the reducing reagent and HNO₃ is the oxidising agent.
oxidation half reaction
Cu ---> Cu²⁺ + 2e --1)
reduction half reaction
4H⁺ + 3e + NO₃⁻ ---> NO + 2H₂O --2)
to balance the number of electrons , 1) x3 and 2) x2
3Cu ---> 3Cu²⁺ + 6e
8H⁺ + 6e + 2NO₃⁻ ---> 2NO + 4H₂O
add the 2 equations
3Cu + 8H⁺ + 2NO₃⁻ ---> 3Cu²⁺ + 2NO + 4H₂O
add 6 nitrate ions to both sides to add up to 8 and form acid with 8H⁺ ions
3Cu + 8HNO₃ ---> 3Cu(NO₃)₂ + 2NO + 4H₂O
Balanced equation for the redox reaction is as follows;
3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O<span>(l)
NO has a coefficient of 2
</span>
Answer: The concentrations of
at equilibrium is 0.023 M
Explanation:
Moles of
= 
Volume of solution = 1 L
Initial concentration of
= 
The given balanced equilibrium reaction is,

Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of
at equilibrium is 0.023 M
Answer:
2.1 atm
Explanation:
We are given the following variables to work with:
Initial pressure (P1): 2.5 atm
Initial temperature (T1): 320 K
Final temperature (T2): 273 K
Constant volume: 7.0 L
We are asked to find the final pressure (P2). Since volume is constant, we want to choose a gas law equation that relates initial pressure and temperature to final pressure and temperature. Gay-Lussac's law does this:

We can rearrange the law algebraically to solve for
.

Substitute your known variables and solve:
