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mina [271]
3 years ago
9

What happens when the temperature or pressure of a reaction system at equilibrium is altered? The equilibrium state is unable to

reestablish. The position of equilibrium is maintained. The system will try and offset the change. The system will soon reach a state of chaos.
Chemistry
1 answer:
alina1380 [7]3 years ago
3 0

Answer: The system will try and offset the change.

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 a direction to minimize the effect.

Thus if temperature is increased, the reaction will shift in a direction where temperature is decreasing and vice versa. Similarly if pressure is increased, the reaction will shift in a direction where pressure is decreasing and vice versa.

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In the reaction between the strong acid HCI and the strong base NaOH which of the ions listed below is a spectator ion?
qwelly [4]

Answer:

Na+

Explanation:

The equation would be:

HCl (aq) + NaOH (aq) --> HOH (l) + NaCl (aq)

The equation is already balanced and the NaCl will disassociate in Na+ and Cl- and HCl will disassociate into H+ and Cl- and NaOH will disassociate into Na+ and OH-. Na+ is on both sides of the equation and stays the same, so Na+ will be the spectator ion.  

3 0
3 years ago
Read 2 more answers
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change
MrRa [10]

Answer:

-54 kJ/mol

Explanation:

Given that:

A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol

i.e

50 ml of 1.0 M HCl +  50 ml of 1.0 M NaOH -----> -54 kJ/mol

If he repeat the same experiment with :

100 ml of 1.0 M HCl + 100 ml of 1.0 M NaOH. ------> ????

From The experiment; the molar enthalpy of change of the acid-base neutralization reaction will be -54 kJ/mol

This is because : The second reaction requires 50 ml in order to neutralize the reaction, then the remaining 50 ml will be excess, Hence, there is no change in the enthalpy of the reaction.

Similarly; we can assume that :

In the first reaction;  P moles of  is used to liberate Q kJ heat ; then  the change in molar enthalpy will be Q/P (kJ/mol).

SO; when he used 100 ml ;

then the amount of moles used is double, likewise the heat liberated will be doubled ;

So;

2P moles is used to liberate 2Q kJ heat ;

2P/2Q = Q/P ( kJ/mol) = -54 kJ/mol

8 0
3 years ago
Use the average volume of HNO3 used to calculate the concentration of<br> NaOH ?
inessss [21]

Answer:

Incomplete question. You didn't give the volume of HNO3

8 0
3 years ago
how many significant figures does the following number contain : 0.0972how many significant figures does the following number co
jekas [21]

Answer:

3 significant figures that is 9,7 and 2

8 0
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The following image represents the movement of a tsunami, a giant wave produced by an earthquake at sea. Which type of model is
artcher [175]

Answer:

A

Explanation:

It is not c or b, D is not applicable so it must be A

8 0
3 years ago
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