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VARVARA [1.3K]
3 years ago
14

According to the law of conservation of matter, atoms cannot be created or destroyed during a chemical reaction. what do you thi

nk happened to the aluminum atoms after the reaction?
Chemistry
1 answer:
Zarrin [17]3 years ago
3 0

The word atom comes from the Greek atomos which means indivisible, indivisible (excepted in the nuclear reactions). It was therefore perfectly suited to the concept of atom when it was developed by chemists in the early nineteenth century, ie as the smallest part of a pure body.

So the atom of aluminum will remain unchanged, whatever the reaction it will undergo, it is only its molecular form that will change.

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How many mole of 88 gram of water
AnnZ [28]
Mole has about 48gram
6 0
3 years ago
Read 2 more answers
The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th
OLga [1]

Answer:

ΔH = -162.5 kJ.

Explanation:

Hello.

In this case, we first rearrange the reactions:

ClO(g) + O₃(g) ⇒ Cl(g) + 2O₂(g);  ΔH =-122.8 kJ

2O₃(g) ⇒ 3O₂(g);  ΔH=-285.3 kJ

O₃(g) + Cl(g) ⇒ ClO(g) + O₂(g);  ΔH= ?

Thus, we are going to use the Hess law, as an strategy to rearrange the known chemical reactions and thereby compute the enthalpy of reaction of the unknown one.

1. The first reaction must be inverted in order to obtain chlorine as a reactant in the third one, therefore, the enthalpy of reaction becomes positive:

Cl(g) + 2O₂(g) ⇒ ClO(g) + O₃(g);   ΔH = 122.8 kJ

2. Second reaction remains the same:

2O₃(g) ⇒ 3O₂(g);  ΔH=-285.3 kJ

Then, we add them to obtain:

Cl(g) + 2O₂(g) + 2O₃(g) ⇒ ClO(g) + O₃(g) + 3O₂(g)

Whereas we can subtract both oxygen and ozone to obtain the third one:

O₃(g) + Cl(g) ⇒ ClO(g) + O₂(g)

Therefore, the enthalpy of reaction turns out:

ΔH = 122.8 kJ + (-285.3 kJ )

ΔH = -162.5 kJ.

Best regards.

4 0
3 years ago
Be sure to answer all parts. Consider the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g)ΔH = −198.2 kJ/mol How would the concentrations of S
andrew-mc [135]

Answer:

a)

[SO2]: The concentration increases

[O2]: The concentration increases.

[SO3]: The concentration decreases.

b)

[SO2]: The concentration decreases.

[O2]: The concentration decreases.

[SO3]: The concentration increases.

c)

[SO2]:There is no change.

[O2]: There is no change.

[SO3]: There is no change

Explanation:

For an exothermic reaction, increase in temperature decreases the concentration of products and increases the concentration of reactants since increase in temperature shifts the equilibrium position to the left hand side.

Increase in pressure and decrease in volume will shift the equilibrium position towards the right hand side which means more SO3 in the system.

Catalyst increases the rate of forward and reverse reaction simultaneously hence at equilibrium, the concentration of reactants and products remain unchanged.

4 0
3 years ago
11. Which one of the following is a correct formation reaction?
gavmur [86]
From the choices presented above, I can say the correct answer is option D. The reaction that falls under the formation reaction type is <span>6C(graphite) + 6H2O(s) → C6H12O6(s). This is because the reactants combine and form only one single large product.</span>
7 0
3 years ago
Calculate the freezing point of a solution made from 52.6 g of propane, c3H8, dissolved in 196.0 g of benzene, C6H6. The freezin
OLga [1]
Given: weight of solute (propane) = 52.6 g
weight of solvent (benzene) = 196 g = 0.196 kg
We know that, molecular weight of propane = 44.1 g/mol

∴ Molality of solution = \frac{\text{weight of solute (g)}}{\text{Molecular weight X Weight of solvent (Kg)}}
                                  =  \frac{52.6}{44.1X.196} = 6.085 m

Now, Depression of freezing point = Kf m
where Kf = cryoscopic constant = <span>5.12 oC/m

</span>∴  Depression of freezing point = 5.12 X 6.085
                                                  = 31.15 oC

Therefore, freezing point of solution = 5.50 - 31.15 = -25.65 oC
4 0
3 years ago
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