Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
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2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
Special Structures in Plant Cells. Most organelles are common to both animal and plant cells. However, plant cells also have features that animal cells do not have: a cell wall, a large central vacuole, and plastids such as chloroplasts.
Answer:
Explanation:
The breakdown reaction of ozone is as follows
It can be seen that 2 moles of ozone is required in the complete cycle
So for 10 cycles, 20 moles of ozone is required
m = Mass of 
= 15.5 g
M = Molar mass of = 104.46 g/mol
P = Pressure = 24.5 mmHg
T = Temperature = 232 K
R = Gas constant = 
Number of moles is given by
From ideal gas law we have
For 20 cycles of the reaction the volume of the ozone is .
• a fuel gas consisting mainly of carbon monoxide and hydrogen, made by passing steam over incandescent coke
• the cranium and the mandible.
hope it helps...!!!
Answer:
ΔH°r = -1562 kJ
Explanation:
Let's consider the following combustion.
C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)
We can calculate the standard heat of reaction (ΔH°r) using the following expression:
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(i) are the standard heats of formation of reactants and products
The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]
ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]
ΔH°r = -1562 kJ
