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icang [17]
3 years ago
9

Which example is an endothermic process

Chemistry
1 answer:
Annette [7]3 years ago
4 0

Answer:

Endothermic reactions is the result of heat being absorbed.

Explanation:

Melting ice cubes. Melting solid salts. Evaporating liquid water. Converting frost to water vapor (melting, boiling, and evaporation, in general, are endothermic processes.

(These are all examples of endothermic reactions.)

You might be interested in
A chemical equilibrium between gaseous reactants and products is shown.
fenix001 [56]

Explanation:

Question 1:

It will shift toward the reactant side as there are a greater number of moles of gas on the reactant side.

Pressure changes only affects only equilibrium involving a gas or gases. An increase in pressure shifts the equilibrium in the direction of the reaction with lower volume.

                 N₂  + 3H₂   ⇆   2NH₃

                        4                  2

An increase in pressure will shift the equilibrium towards the product side.

If the pressure is decreased, equilibrium will shifts towards the side of the reactants. This is because the volume of the reactants are higher than that of the product.

Question 2:

True

Based on Le Chatelier's principle, an increase in temperature will shift the equilibrium position towards the products in an endothermic reaction.

An endothermic reaction is a reaction that absorbs heat from the surroundings.

A rise in temperature shifts equilibrium position to the direction that absorbs the heat and vice versa.

In an endothermic reaction, the forward reaction absorbs the heat and therefore equilibrium shifts towards the side of product.

Question 3:

Decreasing the pressure              A) Shift to the left

Adding hydrogen gas                   B) Shift to the right

Adding a catalyst                          C) No effect

                N₂  + 3H₂   ⇆   2NH₃

Pressure : Decreasing the pressure will shift the equilibrium position to the left. In reactions involving gases, increase in pressure favors the side with a lesser volume. An increase in pressure will favor the formation of ammonia as a product.

A decrease in pressure shifts the equilibrium to the left towards the side of high volume.

Concentration: addition of hydrogen gas will increase the concentration of the gas. An increase in concentration favors the side that uses up the specie and lowers it concentration.

If the concentration of hydrogen gas is increased the equilibrium will shift to the right to annul the concentration of reactants that has increased.

Catalyst: Catalysts have no effect on the position of chemical equilibrium. It only affects the rate at which equilibrium can be reached.

Question 4:

Adding more of gas C to the system

The change that would shift the equilibrium system to the left is by adding more of gas C to the system.

This will increase the concentration of gas C, in order to annul this, the equilibrium will shift to the left.

  • Heating the system will shift the equilibrium to the right
  • Increasing volume only affects purely gaseous reactions.
  • Removing some of gas C from the system shifts the equilibrium to the right.

Question 5:

Heating the system

This is an exothermic reaction because enthalpy change is negative. A rise in temperature shifts equilibrium position to the direction that absorbs heat.

In an exothermic reaction, heat is given off in the forward process. The reverse process is endothermic and heat is absorbed.

Therefore, heating the system will shift the equilibrium to the left and heat is absorbed.

Question 6:

This reaction is exothermic because the system shifted to the left on heating.

It is right to conclude that this reaction is exothermic because the system shifted to the left on heating .

Since the reaction turned dark brown on heating, it shows that more of the reactant Nitrogen dioxide gas was produced.

We can see that the reaction is exothermic in the forward process. An increase in temperature shifts equilibrium position backward to the side that absorbs the heat. This allows for the production of more nitrogen dioxide.

Question 7:

Shift it toward the reactants

A decrease in the concentration of the reactants shift the reaction towards the sides of the reactants.

An increase in concentration of a specie favors the direction that uses up that specie and lowers its concentration.

If the concentration of a specie on the reactant side is increased, equilibrium shifts to the right where more products are formed.

if we decrease the concentration of reactants, the products will be more concentrated and the equilibrium will shift to the left i.e the reactant side.

Learn more:

Equilibrium constant brainly.com/question/11126965

#learnwithBrainly

3 0
4 years ago
A solution that contains a small amount of salt in a large amount of water is said to be a -----solution?
Tems11 [23]
It's a saline solution, commonly used in many ways, primarly in chemistry and medicine (i.e. in intravenous therapy, nasal lavage or in biological and chemical experiments), but also for cleaning new piercings or rinsing contact lenses.
5 0
3 years ago
an alloy composed of tin, lead, and cadmium is analyzed. the mole ratio of sn:pb is 2.73:1.00, and the mass ratio of pb:cd is 1.
nordsb [41]

This problem is describe the mole-ratio composition of an allow composed by tin, lead and cadmium. Ratios are given as Sn:Pb 2.73:1.00 and Pb:Cd is 1.78:1.00, and we are asked to calculate the mass percent compositon of Pb in the allow.

In this case, according to the given information, it turns out possible realize that the following number of moles are present in the alloy, according to the aforementioned ratios:

2.73mol Sn\\\\1.00molPb\\\\\frac{1.00molPb*1.00molCd}{1.78molPb}= 0.562molCd

Next, we calculate the masses by using each metal's atomic mass:

m_{Sn}=2.73mol*\frac{118.7g}{1mol}=324.05g\\\\ m_{Pb}=1.00mol*\frac{207.2g}{1mol}=207.2g\\\\m_{Cd}=0.562mol*\frac{112.4g}{1mol}=63.2g

Thus, the mass percent composition of each metal is shown below:

\%Sn=\frac{324.05g}{324.05g+207.2g+63.2g} *100\%=54.5\%\\\\\%Pb=\frac{207.2g}{324.05g+207.2g+63.2g} *100\%=34.9\%\\\\\%Cd=\frac{63.2}{324.05g+207.2g+63.2g} *100\%=10.6\%

So that of lead is 34.9 %.

  • brainly.com/question/19168996
  • brainly.com/question/2578569

Learn more:

8 0
3 years ago
There is a part A and B
ladessa [460]

Answer:

Im so confused, and maybe blind bcs I cant read it

Explanation:

Could you explain??

3 0
3 years ago
Read 2 more answers
For the process 2SO2(g) + O2(g) --> 2SO3(g),
CaHeK987 [17]

Answer:

–187.9 J/K

Explanation:

The equation that relates the three quantities is:

\Delta G = \Delta H - T \Delta S

where

\Delta G is the Gibbs free energy

\Delta H is the change in enthalpy of the reaction

T is the absolute temperature

\Delta S is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K

8 0
4 years ago
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