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3 years ago

What types of reactions are these 3 chemical equations?-

Chemistry

1 answer:

IceJOKER [234]3 years ago

3 0

Answer:

*2Kl+Pb(NO3)2=PbI2+2KNO3: double replacement.

*2Al+3CuSO4=Al2(SO4)3+3Cu: single replacement.

*C2H5OH+3O2=2CO2+3H2O: combustion.

Explanation:

Hello there!

In this case, according to the required, it turns out necessary for us to recall the five types of reactions, combination, decomposition, single and double replacement and combustion as shown on the attached figure.

In such a way, since the first reaction follows the pattern AB+CD-->AD+CB we infer it is double replacement; the second reaction follows the patter A+BC-->AC+B and therefore it is single replacement; and the last one follows the pattern of combustion reaction due to the presence of CO2 and H2O on the products side.

Regards!

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The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for

tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. $K(s)+O_2(g)\rightarrow KO_2(s)$

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$

Answer: A. $K(s)+O_2(g)\rightarrow KO_2(s)$ : decreases

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g)$ : decreases

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ : no change

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction $K(s)+O_2(g)\rightarrow KO_2(s)$ , as gaseous reactant is changed to solid product, entropy decreases.

In reaction $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0

3 years ago

Which statement best describes a solution?

lara31 [8.8K]

A mixture having a uniform composition where the components can't be seen separately and all components are in the same state best describes a solution.

<span>In chemistry, a </span>solution<span> is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.</span>

The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago

Read 2 more answers

Aluminum has a face-centered cubic unit cell. How many atoms of Al are present in each unit cell

Serjik [45]

Answer:4

Explanation:

As shown in the image attached, a face-centred cubic structure has 8 atoms at the corners and 6 face center atoms.

Each corner atom contributes to eight cell, so per unit cell 1/8 ×8 =1atom

Face center atoms contributes to two unit cells 1/2 × 6=3atoms

Total atoms =3+1=4atoms

Therefore the atoms in Al FCC per unit cell is 4

7 0

3 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$