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pantera1 [17]
4 years ago
10

If temperature increases from 65 degrees to 90 degrees, does the solubility increase or decrease for KBr during the same interva

l and by how much?

Chemistry
1 answer:
maria [59]4 years ago
5 0

If the heat given off in the dissolving reaction is less than the heat required to break apart the solid, the net dissolving reaction is endothermic. The addition of more heat facilitates the dissolving reaction by providing energy to break bonds in the solid. This is the most common situation where an increase in temperature produces an increase in solubility for solids.

The use of first-aid instant cold packs is an application of this solubility principle. A salt such as ammonium nitrate is dissolved in water after a sharp blow breaks the containers for each. The dissolving reaction is endothermic - requires heat. Therefore the heat is drawn from the surroundings, the pack feels cold.

The effect of temperature on solubility can be explained on the basis of Le Chatelier's Principle. Le Chatelier's Principle states that if a stress (for example, heat, pressure, concentration of one reactant) is applied to an equilibrium, the system will adjust, if possible, to minimize the effect of the stress.  This principle is of value in predicting how much a system will respond to a change in external conditions.  Consider the case where the solubility process is endothermic (heat added). An increase in temperature puts a stress on the equilibrium condition and causes it to shift to the right.  The stress is relieved because the dissolving process consumes some of the heat. Therefore,  the  solubility  (concentration)  increases  with  an  increase  in  temperature.    If  the process is exothermic (heat given off). A temperature rise will decrease the solubility by shifting the equilibrium to the left.

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Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets:
Tcecarenko [31]

Answer:

1.25~mol~H_2O and 0.627~mol~N_2

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine (N_2H_4) and <u>oxygen</u> (O_2). So, we can start with the <u>reaction</u> between these compounds:

N_2H_4~+~O_2~->~N_2~+~H_2O

Now we can <u>balance the reaction</u>:

N_2H_4~+~O_2~->~N_2~+~2H_2O

In the problem, we have the values for both reagents. Therefore we have to <u>calculate the limiting reagent</u>. Our first step, is to calculate the moles of each compound using the <u>molar masses values</u> (32.04 g/mol for N_2H_4 and 31.99 g/mol for O_2):

20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4

20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2

In the balanced reaction we have 1 mol for each reagent (the numbers in front of O_2 and N_2H_4 are 1). Therefore the <u>smallest value would be the limiting reagent</u>, in this case, the limiting reagent is N_2H_4.

With this in mind, we can calculate the number of moles for each product. In the case of N_2 we have a <u>1:1 molar ratio</u> (1 mol of N_2 is produced by 1 mol of N_2H_4), so:

0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2

We can follow the same logic for the other compound. In the case of H_2O we have a <u>1:2 molar ratio</u> (2 mol of H_2O is produced by 1 mol of N_2H_4), so:

0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O

I hope it helps!

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