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
Its an ore of uraninte i think.
Answer:
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Explanation:
Let volume of the 40% acid solution be x.
Let volume of the 60% acid solution be y.
Volume of solution formed after mixing both solution = 40 L
x + y = 40 L..[1]
Volume of acid 40% solution = 40% of x= 0.4x
Volume of acid 60% solution = 60% of y= 0.6y
Volume of acid formed = 45% of 40 L = 
..[2]
Solving [1] and [2]
x = 30 L , y = 10 L
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Answer:u would have 40
Explanation:
because ur taking them away
Answer: 2.88×
atoms 
Explanation: First, using stoichiometry, we must convert this from grams to moles, then from moles to atoms.
1. For the first step, we should also look at the periodic table to find the molar mass of the compound, then use that as the denominator.
2. Now that it is converted to moles, we must convert it to atoms by multiplying it by Avogadro's number.
With this information, we know that there are 
total atoms in 0.680 grams 
.
I hope this helps! Pls give brainliest!! :)
