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riadik2000 [5.3K]
2 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:
swat322 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]2 years ago
5 0

Answer:

d because i got a 100%

Explanation:

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6 0
2 years ago
3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g
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The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g  of a dibasic acid

Required

the relative molecular mass of acid A

Solution

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M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

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mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

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3 years ago
Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and
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Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts.  Since the stoichiometry is 1:1, x amount of O₂ also reacts.  This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

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Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x.  So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0
3 years ago
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