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3 years ago

How many KJ are given off when 120 grams of water are cooled from 25°C to -25°C?

Chemistry

1 answer:

Pavel [41]3 years ago

4 0

-25.116 Kilojoules are given off when 120 grams of water are cooled from 25°C to -25°C.

Explanation:

GIVEN THAT:

mass of water = 120 grams OR 0.12 Kg

initial temperature = 25 degrees

final temperature = -25degrees

change in temperature (ΔT) = final temperature - initial temperature

-25 -25

= -50 degrees

specific heat capacity of water = 4.186 joule/gram degree Celsius

Q (heat energy transferred) = ?

Formula used,

Q = mcΔT

Putting the values in equation

Q = 120 X 4.186 X -50

= -25116 joules

The energy in joules is converted to kilo joule by dividing it with 1000.

The negative sign in the energy transfer shows that energy is released in the process.

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Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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