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3 years ago

What is Artificial selection???

Chemistry

1 answer:

kati45 [8]3 years ago

7 0

Answer:

The process of selection conducted under human direction.

Explanation:

For example, by allowing only like individuals to breed, breeders have created the great variety of dog breeds and crop plants

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How many pairs of electrons are shared between the nitrogen atoms in a molecule of n2?

Kobotan [32]

An N∨2 Molecule shares 3 pairs of electrons and forms a triple bond. Each N atom contributes 3 electrons making three pairs of electrons.

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4 years ago

What is the mass of 0.595 mol of the element potassium (K)?Answer in units of g.

oksano4ka [1.4K]

Answer:

2.33 *10^1 g

Explanation:

.595 mol *39.0983 g/1 mol = 23.2634 > 2.33 *10^1 g

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3 years ago

HELP PLZ!!! A sample contains only the compound CaCl2 and water. When it is dried, its weight decreases from 5.00g to 4.24g. Wha

steposvetlana [31]

Answer:

Mass of water is 18.01528 g/mol. I hope this helps

Explanation:

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3 years ago

Read 2 more answers

What is the volume of a substance with a mass of 64.9 g/ and a density of 2.1 g/ml?

maks197457 [2]

It’s 34 I had this question

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3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago