Answer:
0.0277 M.
Explanation:
The integral rate law of a first order reaction:
<em>Kt = ln ([A₀]/[A]),</em>
where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,
t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,
[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>
<em>∵ Kt = ln ([A₀]/[A]),</em>
∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]
Taking the exponential of both sides:
1.6 = (0.0445 M)/[A]
<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>
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At STP one mol weighs 22.4L
Moles of O_2
1 mol.O_2 can create 2mol water
moles of water
Volume of water
Answer:
a) equilibrium shifts towards the right
b) equilibrium shifts towards the right
c) equilibrium shifts towards the left
d) has no effect on equilibrium position
e) has no effect on equilibrium position
Explanation:
A reversible reaction may attain equilibrium in a closed system. A chemical system is said to be in a state of dynamic equilibrium when the rate of forward reaction is equal to the rate of reaction.
According to Le Chateliers principle, when a constraint such as a change in temperature, pressure, volume or concentration is imposed upon a system in equilibrium, the equilibrium position shifts in such a way as to annul the constraint.
When the concentration of reactants is increased, the equilibrium position is shifted towards the right hand side and more products are formed. For an endothermic reaction, the reverse reaction is favoured by a decrease in temperature. Increase in pressure has no effect on the system since there are equal volumes on both sides of the reaction equation. Similarly, the addition of a catalyst has no effect on the equilibrium position since it speeds up both the forward and reverse reactions to the same extent.
<span>oxidizing substance removes electrons from another substance, which are then added to itself, the oxidizing substance becomes “reduced” (more negative). And because it “accepts” electrons .</span>
