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3 years ago

A population of animals has a short gestational period and travels in large groups. How will this affect the population size?

Chemistry

1 answer:

Katen [24]3 years ago

6 0

It will affect the population size through the following:

High birth rate

Increase in individuals

<h3></h3><h3>What is Gestational period?</h3>

Gestational period is the period in which organisms carries its young in the

uterus before delivery. Animals which have short gestational period have a

high birth rate as they can get pregnant as many times a possible during the

year.

This therefore brings about an increase in the individuals present in the

ecosystem. The option A is therefore the most appropriate choice.

Read more about Gestational period here brainly.com/question/1305202

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How many total moles of reactants and how many total moles of products are in the reaction 2NO, (g) +

zavuch27 [327]

Answer:

moles reactant = 2, moles of products = 3

Explanation:

The reactants are on the left side of the equation. Although, energy (heat) is a reactant it will not factor into the calculation for moles. The coefficient is the number of moles for each substances. So for the reactant side NO would have 2 moles because the coefficient is 2. Using the same logic, NO on the product side will have 2 moles and O2 will be 1 mole. O2 has one mole because it is implied that you know anything multipled by one is the same number. So 1 mole of O2 is written as O2 and not 1 O2.

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2 years ago

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3 years ago

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The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

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3 years ago

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4 years ago