Ask question

Ask question

All categories

3 years ago

14

Calculate the molarity of a solution prepared by dissolving 11.9 g of beryllium chloride in a total volume of 292.2 mL of soluti

on.

1 answer:

anyanavicka [17]3 years ago

8 0

Answer:

8793

Explanation:

You might be interested in

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago

Name the three parts of the atom.

mart [117]

A. Protons, neutrons, and electrons

6 0

3 years ago

When an atom _______<br> an electron, it charge is negative .

nikitadnepr [17]

Answer:

Gains

Explanation:

If it gains an electron it causes the atom to become negative since electrons are negative and an overall charge of an atom is neutral. Therefore, it would become negative.

Hope this helps :)

3 0

2 years ago

Read 2 more answers

What factor affects the solubility of all solutes, whether solid, liquid, and gaseous? -pressure -temperature -volume -number of

Ratling [72]

Answer:

temperature

Explanation:

The solubility is the maximum amount of the substance that will be dissolved in water given the amount of the solvent at a specif temperatures and thus there are factors in which the solubility depends as the temperatures.

The temperature affects the solids and liquids and pressure the gases. The <u>solubility </u>of the <u>organic compounds also increases with temperature changes.</u>

4 0

3 years ago