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Kaylis [27]
3 years ago
12

What is the element above the element in Group 3, Period 4?

Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
5 0
Scandium? Is that what you mean?
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The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product
Darya [45]

Answer: It takes 3.120 seconds for the concentration of  A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

k = rate constant = 0.610M^{-1}s^{-1}

a_0 = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}

t=3.120s

Thus it takes 3.120 seconds for the concentration of  A to decrease from 0.860 M to 0.260 M.

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4 years ago
How many moles of KCl are required to prepare 1.50 L of 5.4 M KCl?
puteri [66]
5.4 M = moles of solute / 1.50 L 

<span>Multiply both sides by 1.50 L to isolate moles of solute on the right. </span>

<span>8.1 mol = moles of solute </span>


6 0
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