This is an incomplete question, here is a complete question.
The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?
Answer : The time taken will be, 17.0 hr
Explanation :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = 
t = time passed by the sample = ?
a = initial concentration of the reactant = 0.080 M
a - x = concentration left = 0.053 M
Now put all the given values in above equation, we get
Therefore, the time taken will be, 17.0 hr
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
<u>Answer:</u> 1.0 kilograms.
<u>Explanation:</u>
One kilogram is equal to a thousand grams.
Supposing x to be the number of kilograms equal to one thousand and eight grams, we can write it as:
1 kg = 1000 grams
x kg = 1008 grams
To solve for x, we can simply divide 1008 grams by 1000 thousand grams to get the answer.
x = 1008 / 1000
x = 1.008
Rounding this value to the nearest tenth, it will become 1.0 kilograms.
Answer: the theory that all matter us made up of tiny invisible particles. Hope that help you out :)
Explanation:
