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
The missing part of the equation is found to be 4/2He. Option A
<h3>What are nuclear equations?</h3>
The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.
Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.
Learn more about nuclear equations:brainly.com/question/19752321
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The alloy has a density of 21.186g/cc. So for a kilogram or 1000 grams/21.186 g/cc= 45.7 cc. So the answer is 45.7 cc of the allow to make up a kilogram which shows that the density of the allow can be used to calculate the volume of a larger mass ie the kilogram.
Answer:
The equilibrium will shift to produce less ammonia
Explanation:
According to Le Chatelier's principle, the reaction will try to oppose anything that is done on it, if it was at equilibrium.
When the concentration of H2 is decreased, you are decreasing the concentration of hydrogen so the reaction tries to increase the concentration of hydrogen by breaking down the ammonia on the products side. This will decrease the output of ammonia
