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2 years ago

15

A solution has a concentration of 0.6mol/dm³. If a container of this solution holds 3 moles of solute, what volume of solution i

s in the container?

1 answer:

FinnZ [79.3K]2 years ago

6 0

Answer: 3 moles solute x 1 dm^3/0.60 moles solute = 5 dm^3

Explanation:

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Consider the dissolution of 1.50 grams of salt XY in 75.0 mL of water within a calorimeter. The temperature of the water decreas

-BARSIC- [3]

Answer:

The quantity of heat lost by the surroundings is 258,5J

Explanation:

The dissolution of salt XY is endothermic because the water temperature decreased.

The total heat consumed by the dissolution process is:

4,184 J/g°C × (75,0 + 1,50 g) × 0,93°C = 297,7 J

This heat is consumed by the calorimeter and by the surroundings.

The heat consumed by the calorimeter is:

42,2 J/°C × (0,93°C) = 39,2 J

That means that the quantity of heat lost by the surroundings is:

297,7J - 39,2J = <em>258,5 J</em>

I hope it helps!

8 0

2 years ago

For the reaction PCl5(g) <--> PCl3(g) Cl2(g) at equilibrium, which statement correctly describes the effects of increasing

xenn [34]

The given question is incomplete. The complete question is :

For the reaction $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$ at equilibrium, which statement correctly describes the effects of increasing pressure and adding $PCl_5$ , respectively

a) Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

b) Increasing pressure causes shift to products ,adding $PCl_5$ causes shift to reactants.

c) Increasing pressure causes shift to products, adding $PCl_5$ causes shift to products.

d) Increasing pressure causes shift to reactants,adding $PCl_5$ causes shift to reactants

Answer: Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

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

a) If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.

b) If $PCl_5$ is added, the equilibrium will shift in the direction where $PCl_5$ is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.

8 0

2 years ago

What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?

Anna11 [10]

Molar mass CaCl₂ = 110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 = 0.7686 / V

V = 0.7686 / 0.788

V = 0.975 L

hope this helps!

5 0

2 years ago

Select the true statements about denaturation.

fiasKO [112]

Answer:

The correct options are: Denaturation of an enzyme will cause a loss of its catalytic activity.

Egg white meringue contains denatured egg white proteins.

A denatured protein has a different tertiary structure than its native state.

Protein denaturation involves cleavage of its peptide bonds.

Explanation:

Protein denaturation is the breakdown of tertiary or quaternary junctions of proteins, which are the easiest to break compared to primary and secondary.

Denaturation can arise due to a change in pH or due to excess temperature. An enzyme is generally a protein and if it is subjected to acidic acids or high temperatures, it denatures, ceasing to perform its function.

Zymogen is a non-denatured INACTIVE proenzyme, that is why we consider that option false along with those that were not selected.

5 0

2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

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

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

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

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago