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2 years ago

When 500 g of water is cooled from 80.0°C to 10.0°C, how much heat energy is lost?

Chemistry

2 answers:

vampirchik [111]2 years ago

6 0

I don’t know

Explanation:

Rashid [163]2 years ago

4 0

This is a solved problem that is close I cannot tell u the direct answer but I hope this STEP BY STEP layout will help

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The main function of the respiratory system is to –

34kurt

The answer is d I believe

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2 years ago

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Explain why it is harder to remove an inner shell electrons than a valence electron from an atom

VMariaS [17]

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3 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

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

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

2 years ago

In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

morpeh [17]

Answer:

The time taken for the cross to become invisible decreases.

Explanation:

We know that one of the factors affecting the rate of reaction is the concentration of reactants. From the collision theory, we know that the higher the concentration of reactants, the greater the possibility of effective collision between reactants leading ultimately to an increase in the rate of reaction. Increase in the rate of reaction implies that the reaction takes a shorter time to reach completion.

In the case of the reaction shown in the question, the point when the reaction is completed is observed by the time take for the cross mark to become invisible. If we look at the given data closely, we will notice that the volume of acid was held constant, the volume of thiosulphate was increased gradually while the volume of water was decreased accordingly. This implies that the concentration of the reactants was increased. Decreasing the volume of water increases reactant concentration.

As explained above, increase in reactant concentration increases the rate of reaction. Hence, the rate of reaction of the acid and thiosulphate increases as reactant concentration increases and the cross mark becomes invisible faster. This implies that in the last column for time taken for the cross to become invisible, the values of time decreases steadily as concentration of reactants increases.

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3 years ago

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2 years ago

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