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3 years ago

What happens to the size of a population if the birth rate decreases while the death rate remains the same?

Chemistry

1 answer:

Sever21 [200]3 years ago

3 0

Answer: It will decrease.

Explanation:

If you had an equal birth and death rate, the population would theoretically stay the same, but since you have more deaths and less births, the population will decrease.

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What are some common examples of solids, liquids and gas?

Sonja [21]

Solids- wood, sand, brick, rock

Liquids- water, milk, blood, coffee

Gases- air, helium, nitrogen, hydrogen

3 0

3 years ago

Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)

MrRissso [65]

Answer:

Equilibrium constant expression for $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)$ :

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}$ .

Where

$a_{\mathrm{H_2CO_3}}$ , $a_{\mathrm{H^{+}}}$ , and $a_{\mathrm{CO_3}^{2-}}$ denote the activities of the three species, and
$[\mathrm{H_2CO_3}]$ , $\left[\mathrm{H^{+}}\right]$ , and $\left[\mathrm{CO_3}^{2-}\right]$ denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator. $\rm H_2CO_3\; (aq)$ is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be $\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)$ .

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient " $2$ " on the product side of this reaction. $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ is equivalent to $\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ . The species $\rm H^{+}\, (aq)$ appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

$\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right$ .

That's where the exponent " $2$ " in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}$ .

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous " $(\rm aq)$ " species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}$ .

8 0

3 years ago

Chlorofluorocarbons such as CCl2F and CCI,F, have been linked to ozone depletion in Antarctica. In 1994, these gases were found

Ilya [14]

Answer:

a) molarity of CCl3F = 1.12 × 10^-11 mol/dm³

Molarity of CCl2F2 = 2.20 × 10^-11 mol/dm³

B) molarity of CCL3F = 7.96 × 10 ^-13 mol/dm³

Molarity of CCl2F2 = 1.55 × 10^-12 mol/dm³

Explanation:

Using the ideal gas equation:

PV = nRT

Further explanations are found in the attachment below.

4 0

4 years ago

Lithium and nitrogen react to produce lithium nitride:

garri49 [273]

Answer:

B : 0.133 M

Explanation:

moles Li3N3 = 0.4 mol Li x (2 moles Li3N/6 moles Li) = 0.133 M

7 0

3 years ago

How many atoms of of titanium are present in a sample containing 0.375 moles Ti

Natasha2012 [34]

2.258302785 x $10^{23}$ atoms of titanium are present in a sample containing 0.375 moles of Ti.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

To find the number atoms we need to multiply by 6.02214076 × 10^{23}.

0.375 moles x 6.02214076 × $10^{23}$

2.258302785 x $10^{23}$

Hence, 2.258302785 x $10^{23}$ atoms of titanium are present in a sample containing 0.375 moles of Ti.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0

2 years ago