The actual yield is 43 g Cl₂.
The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.
∴ The <em>theoretical yield</em> is 60.25 g Cl₂.
% yield = actual yield/theoretical yield × 100 %
Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g
I'm taking this lesson now, so imma help u ( if u need anything else ask me)
so given Molar mass= 32 g/mol
molar mass= (empirical formula) n
32 = (14x1 + 2x1) n
32 = 16 n , so n= 2
so, molecular formula= N2H4
V=abc
a = 2,3cm
b=12,2mm = 1,22cm
c = 0,75inch = 1,905cm
V = 2,3cm*1,22cm*1,905cm ≈ 5,35cm³
Answer:
a. -0.63 V
b. No
Explanation:
Step 1: Given data
- Standard reduction potential of the anode (E°red): -1.33 V
- Minimum standard cell potential (E°cell): 0.70 V
Step 2: Calculate the required standard reduction potential of the cathode
The galvanic cell must provide at least 0.70V of electrical power, that is:
E°cell > 0.70 V [1]
We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.
E°cell = E°cat - E°an [2]
If we combine [1] and [2], we get,
E°cat - E°an > 0.70 V
E°cat > 0.70 V + E°an
E°cat > 0.70 V + (-1.33 V)
E°cat > -0.63 V
The minimum E°cat is -0.63 V and there is no maximum E°cat.
