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riadik2000 [5.3K]
3 years ago
9

Which scenario would result in a population growth of zero? Use this formula: population growth = (birth rate + immigration) – (

death rate + emigration) 12 individuals emigrate, 12 individuals die, 12 individuals are born, 0 individuals immigrate 16 individuals emigrate, 3 individuals die, 13 individuals are born, 0 individuals immigrate 8 individuals emigrate, 0 individuals die, 20 individuals immigrate,12 individuals are born 10 individuals emigrate, 4 individuals die, 14 individuals immigrate, 0 individuals are born
Chemistry
2 answers:
swat323 years ago
8 0

Answer:

12 individuals are born 10 individuals emigrate, 4 individuals die, 14 individuals immigrate, 0 individuals are born

Explanation:

From the formula:

Population growth = (birth rate + immigration) – (death rate + emigration)

                       = (b + i) - (d + e)

<em>For the first option, b = 12, i = 0, d = 12, e = 12</em>

population growth = (12+0) - (12+12) = -12

<em>For the second option: b = 13, i = 0, d = 3, e = 16</em>

population growth = (13+0) - (3+16) = -6

<em>For the third option: b = 12, i = 20, d = 0, e = 8</em>

population growth = (12+20) - (0+8) = 24

<em>For the last option: b = 0, i = 14, d = 4, e = 10</em>

population growth = (0+14) - (4+10) = 0

Hence, the last option is the correct option.

Alex787 [66]3 years ago
5 0

Answer:

d because i got a 100%

Explanation:

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Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

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Answer to your query is provided below

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