a scale-model mound made of the same materials that make the real hill
Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al 
is required.
Therefore for 2 moles of Al,
Total 
F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
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There will be a shift towards the reactants
Answer:
A. 6atm
Explanation:
Using pressure law equation:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
T1 = initial temperature (K)
P2 = final pressure (atm)
T2 = final temperature (K)
According to this question;
P1 = 3 atm
P2 = ?
T1 = 120K
T2 = 240K
Using P1/T1 = P2/T2
3/120 = P2/240
Cross multiply
240 × 3 = P2 × 120
720 = 120P2
P2 = 720/120
P2 = 6atm
<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
