What is the volume of 24.0 g of oxygen gas at STP?

Explain why the dissolved component does not settle out of a solution

Novosadov [1.4K]

<span>explain why the dissolved component does not settle out of a solution -
</span><span>Before saturation, there are attractive forces between solute and solvent. after saturation, the capacity for the attractive forces is reached and no more solute can be dissolved</span>

8 0

2 years ago

What is the name of this alkane? two central carbons are bonded to c h 3 at each end, h below, and c h3 above the left carbon an

Kazeer [188]

The name of this alkane is with central carbons are bonded to c h 3 is 2-methylbutane.

<h3>What is alkane?</h3>

Alkanes belong to the family of saturated hydrocarbons with carbon carbon single bond.

For the given alkane;

CH₃ H

CH₃ - C - C - CH₃

H H

Thus, the name of this alkane is with central carbons are bonded to c h 3 is 2-methylbutane.

Learn more about alkane here: brainly.com/question/24270289

#SPJ4

4 0

1 year ago

Characterize EACH of the three given statements as being TRUE or FALSE and then indicate the collective true-false status of the

Aleks04 [339]

Answer:

A = True

B = False

C = True.

I think this is the answer

8 0

2 years ago

CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o

Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

$CO_{(g)}$ + $2H_2O_{(g)}$ ⇄ $CH_3OH_{(g)}$

The ICE table can be represented as:

$CO_{(g)}$ + $2H_2_{(g)}$ ⇄ $CH_3OH_{(g)}$

Initial 0.27 0.49 0.0

Change -x -2x x

Equilibrium 0.27 - x 0.49 -2x x

We can now say that the concentration of $CH_3OH_{(g)}$ at equilibrium is x;

Let's not forget that at equilibrium $CH_3OH_{(g)}$ = 0.11 M

So:

x = [ $CH_3OH_{(g)}$ ] = 0.11 M

[ $CO_{(g)}$ ] = 0.27 - x

[ $CO_{(g)}$ ] = 0.27 - 0.11

[ $CO_{(g)}$ ] = 0.16 M

[ $2H_2_{(g)}$ ] = (0.49 - 2x)

[ $2H_2_{(g)}$ ] = (0.49 - 2(0.11))

[ $2H_2_{(g)}$ ] = 0.49 - 0.22

[ $2H_2_{(g)}$ ] = 0.27 M

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

$K_C = \frac{(0.11)}{(0.16)[(0.27)^2}$

$K_C = 9.4307$

$K_C$ = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0

2 years ago

Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.

baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that $M_{1}V_{1}$ = $M_{2}V_{2}$

Where, $V_{1}$ = total volume

$V_{2}$ = initial volume

Therefore, new concentration of $CH_{3}COOH$ = $\frac{M_{2}V_{2}}{V_{1}}$

= $\frac{60 \times 0.5}{70}$

= 0.43 M

New concentration of NaOH = $\frac{M_{2}V_{2}}{V_{1}}$

= $\frac{10 \times 1.0}{70}$

= 0.14 M

So, the given reaction will be as follows.

$CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O$

Initial: 0.43 0.14 0

Change: -0.14 -0.14 0.14

Equilibrium: 0.29 0 0.14

As it is known that value of $pK_{a}$ = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

pH = $pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

= $4.74 + log \frac{0.14}{0.29}$

= 4.74 + (-0.316)

= 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

6 0

2 years ago