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Alik [6]
3 years ago
14

The reaction below shows a system in equilibrium.

Chemistry
1 answer:
Marrrta [24]3 years ago
4 0

The given question is incomplete, here is a complete question.

The reaction below shows a system in equilibrium.

H_2(g)+I_2(g)+Heat\rightarrow 2HI(g)

How would a decrease in temperature affect this reaction?

A. The rate of formation of the gases would increase.

B. The equilibrium of the reaction would shift to the left.

C. The equilibrium would shift to cause the gases to sublime into solids.

D. The chemicals on the left would quickly form the chemical on the right.

Answer : The correct option is, (B) The equilibrium of the reaction would shift to the left.

Explanation :

The given reaction is endothermic reaction.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

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.

As, heat is getting absorbed during a chemical reaction. This means that temperature is getting increased on the reactant side.

If the temperature in the equilibrium is decreased, the equilibrium will shift in the direction where, temperature is getting increased. Thus, the reaction will shift in left direction that is towards the reactants.

Hence, the correct option is, (B) The equilibrium of the reaction would shift to the left.

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Fill in the coefficients that will balance the following reaction:
Alex17521 [72]

Answer:

To balance a reaction, the amount of reactants must be equal to the amount of products, as stated by the Law of Conservation of Matter. It may help you to keep track of  the number of each element in a list as you try to balance. It's not able to be balanced.

3 0
3 years ago
Read 2 more answers
why is it harder to remove an electron from fluorine than from carbon? to put it another way, why are the outermost electrons of
Verizon [17]

It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.

The energy required to remove an electron from an atom is called ionization energy.

The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.

Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that;  it is harder to remove an electron from fluorine than from carbon.

Learn more: brainly.com/question/16243729

6 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
3 years ago
What is the atomic number, mass number, element symbol, group number, and period number for oxygen?
yawa3891 [41]

Answer:

Atomic no. = 8

Mass no. = 16

Period no. = 2

Group no. = 6

4 0
3 years ago
What happened after the gas syringe was inserted into the flask with the methane gas?
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