The image shown here is a good illustration of which law?

Place the following substances in the following order: soluble, somewhat soluble, and insoluble.

dimulka [17.4K]

C- Sugar is very much soluble (example- putting sugar in tea), sand is definately not soluble (example- a sandy beach) and cornstarch is the last one, so logically it should be somewhat soluble.

8 0

3 years ago

A 4 liter solution of bleach with a concentration of 0.5 moles/L is diluted using an extra 4 Liters of water. What is the final

blondinia [14]

Answer:

0.25 mol/L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 4L

Initial concentration (C1) = 0.5 mol/L

Final volume (V2) = 4 + 4 = 8L

Final concentration (C2) =?

Applying the dilution formula, we can easily find the concentration of the diluted solution as follow:

C1V1 = C2V2

0.5 x 4 = C2 x 8

Divide both side by 8

C2 = (0.5 x 4 )/ 8

C2 = 0.25 mol/L

Therefore the concentration of the diluted solution is 0.25 mol/L

3 0

3 years ago

The indicator used in this redox titration is sulfuric acid

anygoal [31]

Methyle orange is the indicator that is used in sulfuric acid.

4 0

3 years ago

A compound contains 74.2 g Na and 25.8 g O. Determine the Empirical Formula for this compound

Ksenya-84 [330]

Answer:

The empirical formula for the compound is Na2O

Explanation:

Data obtained from the question include:

Sodium (Na) = 74.2g

Oxygen (O) = 25.8g

We can obtain the empirical formula for the compound as follow:

First, divide the above by their individual molar mass as shown below:

Na = 74.2/23 = 3.226

O = 25.8/16 = 1.613

Next, divide the above by the smallest number

Na = 3.226/1.613 = 2

O = 1.613/1.613 = 1

Therefore, the empirical formula is:

Na2O

4 0

3 years ago

Read 2 more answers

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