Answer:
Heating the system
Explanation:
According to the principle of Le Chatelier, for a system at equilibrium, a specific disturbance would make the equilibrium shift toward the direction which minimizes such a disturbance.
Since we wish to shift the equilibrium to the left, this means we wish to increase the concentration of products, as an excess in their concentration would make the products react and produce more reactants in order to lower the excess concentration of products.
Since heat is also a product, an increase in heat would shift the equilibrium toward the left, as this would consume the excess of heat by producing the reactants.
Answer: from the Zn anode to the Cu cathode
Justification:
1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)
2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).
3) Then, you can already tell that electrons go from Zn to Cu.
4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.
So you get that the electrons flow from the anode (Zn) to the cathode (Cu).
Always oxidation occurs at the anode, and reduction occurs at the cathode.
Answer:
Anion
Explanation:
Anion is an atom or group of atoms that are negatively charged.
Cations are positively charged
Anode is a positively charged pole that conducts electricity (remember electrolysis)
Cathode is a negatively charged pole that conducts electricity (remember electrolysis too)
Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001 
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13 
From ideal gas equation P V = m RT
⇒ P = 
Put all the values in above formula we get
⇒ P = 
× 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
