Ask question

Ask question

All categories

3 years ago

11

What compound is always part of an aqueous solution

2 answers:

const2013 [10]3 years ago

8 0

The compound that is always part of an aqueous solution is NaCI(aq).

Anna [14]3 years ago

3 0

First, an aqueous solution is in which the solvent is Water.
Some aqueous solutions: Sodium Chloride NaCl, Sodium Hydroxide NaOH, Pepsi, and many others

Hope this Helps :)

You might be interested in

Describe how you could show by adding aqueous sodium hydroxide and aqueous ammonia that a solution contained zinc ions.

Pavel [41]

Answer:

*result with sodium hydroxide: white ppt of Zn(OH)2

*excess sodium hydroxide: Precipitate dissolves to form a COLOURLESS solution

*Result with Aqueous ammonia: White ppt of Zn(OH)2

*excess aqueous ammonia: Precipitate dissolves to form a COLOURLESS solution

7 0

3 years ago

How does matter change phase?

crimeas [40]

Answer:

Phase changes typically occur when the temperature or pressure of a system is altered. When temperature or pressure increases, molecules interact more with each other. When pressure increases or temperature decreases, it's easier for atoms and molecules to settle into a more rigid structure.

Explanation:

Hope it helps UvU

7 0

3 years ago

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

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

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

When matter it gains energy and changes from a solid to a liquid as the particles move faster and the bonds holding the pattern

andrezito [222]

When you heat up the matter, it gains energy and the bonds are broken. Then it will make solid become liquid

6 0

3 years ago

What do you think about our current forms of recycling? Why?

tankabanditka [31]

It is absurd. Environmentalists prefer that you don’t recycle at all because most if not all of our recyclables are being recycled incorrectly.

8 0

3 years ago