Reaction involved in present electrochemical cell,
At Anode: Zn → Zn^2+ + 2e^2-
At cathode: Zn^2+ + 2e^2- → Zn
Net Reaction: Zn + Zn^2+ ('x' m) → Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2
According to Nernst equation for electrochemical cell,
Ecell = -2.303
![\frac{RT}{nT} log \frac{[Zn^2+]R}{[Zn^2+]L}](https://tex.z-dn.net/?f=%20%5Cfrac%7BRT%7D%7BnT%7D%20log%20%20%5Cfrac%7B%5BZn%5E2%2B%5DR%7D%7B%5BZn%5E2%2B%5DL%7D%20%20)
= 0.014
Given: T =
, F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v
∴ 0.014 = - 2.303
∴ log
=
= -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m
Answer:
Sodium hydrogen carbonate
Answer:
48 grams
Explanation:
The chemical equation for the reaction is the following:
2 H₂ + O₂ → 2 H₂O
That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:
MW(O₂) = 16 g/mol x 2 = 32 g/mol
mass of O₂ = 1 mol x 32 g/mol = 32 g
So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:
3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂
<em>Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.</em>
Answer:
1. nitric acid: sa
2. perchloric acid: sa
3. hydrofluoric acid: wa
Explanation:
A strong acid (sa) is the one that is completely dissociated into ions in water. Conversely, a weak acid (wa) is not completely dissociated in water.
From the options, the strong acids are:
1. nitric acid (HNO₃). It dissociates completely into ions when is dissolved in water, as follows:
HNO₃ → H⁺ + NO₃⁻
2. perchloric acid (HClO₄). It is completely dissociated in water as follows:
HClO₄ → H⁺ + ClO₄⁻
The weak acid is hydrofluoric acid (HF). In water, only a small proportion is dissociated into ions. The proportion of ions formed is given by the equilibrium constant Ka. The dissociation is written by using double arrows:
HF ⇄ H⁺ + F⁻
