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Karo-lina-s [1.5K]

1 year ago

13

Find the rate constantrif the population doubles in 12 days.b.ifp= 200 initially (whent= 0), what is the population whent= 18 da

ys?

1 answer:

frez [133]1 year ago

5 0

The rate constant is 0.058/days and the population after 18 days is 412.38.

<h3>What is rate constant?</h3>

The rate constant of the population is calculated as follows;

N = N₀(2)^(rt)

where;

r is the rate constant
t of double of the population

when the population doubles, t = 12 days

2N₀ = N₀(2)^(12r)

2 = (2)^(12r)

ln(2) = 12r

r = ln(2)/12

r = 0.058/days

<h3>Population size after 18 days</h3>

The population size after 18 days is calculated as follows;

N = 200(2)^(0.058 x 18)

N = 412.38

Thus, the rate constant is 0.058/days and the population after 18 days is 412.38.

Learn more about rate constant here: brainly.com/question/11272059

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julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be $\frac{1876.8}{g}$

Explanation:

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writing equation of motion in vertical direction:

$v_{y}=u_{y}+(-g) t$

$u_{y}= u\sin \phi$

and $v_{y}=0$

therefore $\frac{u\sin \phi }{g} =t$

writing equation of motion in horizontal direction:

$s_{x}=u_{x}t$

$u_{x} = u\cos \phi$

therefore the equation becomes $s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}$

therefore horizontal distance traveled = $\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}$

5 0

3 years ago

what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s

yaroslaw [1]

Answer : The value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = ?

t = time = 17s

$[A_t]$ = final concentration = 0.0981 M

$[A_o]$ = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

$k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}$

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Therefore, the value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

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3 years ago

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Answer:

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$C_{z}$ = 58 -42 = 16

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3 years ago

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