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3 years ago

Unrestricted populations of organisms experience _____.

Chemistry

2 answers:

Alik [6]3 years ago

6 0

Answer:

1. A

2. D

Explanation:

1. If a population is left unrestricted, meaning it is not limited to anything, then the population will continue to increase exponentially due to there being no limits.

2. If a population is limited, and it has reached its <u>carrying capacity</u>, then the essential needs of life are being competed for. Things like food, shelter, and mates are now a limited resource, and hence the option would be "all of the above".

Fudgin [204]3 years ago

5 0

Unrestricted populations of organisms experience Exponential growth.

As a population reaches is carrying capacity there is an increase in competition for All of the above.
Hope I helped.

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3 years ago

Identify factors that affect the ratio when substances combine to form compounds

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3 years ago

include a detailed explanation of what is happening when the purple color is given off in terms of the electrons.

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2 years ago

Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2 Determine which 100.0 gram sample will have

vampirchik [111]

Answer : The sample $H_2O$ has greatest number of oxygen atoms.

The order of number of oxygen atoms from greatest to least will be:

$H_2O>CO_2>C_3H_6O_2>N_2O$

Explanation :

First we have to calculate the moles of $H_2O,N_2O,C_3H_6O_2\text{ and }CO_2$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}$

Molar mass of $H_2O$ = 18 g/mol

$\text{Moles of }H_2O=\frac{100.0g}{18g/mol}=5.55mol$

and,

$\text{Moles of }N_2O=\frac{\text{Mass of }N_2O}{\text{Molar mass of }N_2O}$

Molar mass of $N_2O$ = 44 g/mol

$\text{Moles of }N_2O=\frac{100.0g}{44g/mol}=2.27mol$

and,

$\text{Moles of }C_3H_6O_2=\frac{\text{Mass of }C_3H_6O_2}{\text{Molar mass of }C_3H_6O_2}$

Molar mass of $C_3H_6O_2$ = 74 g/mol

$\text{Moles of }C_3H_6O_2=\frac{100.0g}{74g/mol}=1.35mol$

and,

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mol

$\text{Moles of }CO_2=\frac{100.0g}{44g/mol}=2.27mol$

Now we have to calculate the number of oxygen atoms in $H_2O,N_2O,C_3H_6O_2\text{ and }CO_2$

In $H_2O$ :

As, 1 mole of $H_2O$ contains $1\times 6.023\times 10^{23}$ number of oxygen atoms

So, 5.55 mole of $H_2O$ contains $5.55\times 6.022\times 10^{23}=3.34\times 10^{24}$ number of oxygen atoms

In $N_2O$ :

As, 1 mole of $N_2O$ contains $1\times 6.023\times 10^{23}$ number of oxygen atoms

So, 2.27 mole of $N_2O$ contains $2.27\times 6.022\times 10^{23}=1.37\times 10^{24}$ number of oxygen atoms

In $C_3H_6O_2$ :

As, 1 mole of $C_3H_6O_2$ contains $2\times 6.023\times 10^{23}$ number of oxygen atoms

So, 1.35 mole of $C_3H_6O_2$ contains $2\times 1.35\times 6.022\times 10^{23}=1.62\times 10^{24}$ number of oxygen atoms

In $CO_2$ :

As, 1 mole of $CO_2$ contains $2\times 6.023\times 10^{23}$ number of oxygen atoms

So, 2.27 mole of $CO_2$ contains $2\times 2.27\times 6.022\times 10^{23}=2.73\times 10^{24}$ number of oxygen atoms

The order of number of oxygen atoms from greatest to least will be:

$H_2O>CO_2>C_3H_6O_2>N_2O$

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3 years ago