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ch4aika [34]
3 years ago
13

Calculate the mass of calcium carbide produced if 1770 kJ of heat is absorbed.

Chemistry
1 answer:
Flauer [41]3 years ago
7 0

Answer:

m=243.72g

Explanation:

Hello,

For the given chemical reaction:

CaO(s) + 3 C(s) \rightarrow CaC_2(s) + CO(g)

Whose heat of reaction is 464.8 kJ/mol which means that energy is absorbed due to the chemical reaction, we can compute the moles of calcium carbide via the following relationship:

n=\frac{1770kJ}{464.8 kJ/mol}\\ \\n=3.81mol

Then, since the molar mass of calcium carbide is 64 g/mol, the yielded mass turns out:

m=3.81mol*\frac{64g}{1mol}\\ \\m=243.72g

Best regards.

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Answer: is is true

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Identify the FALSE statement. You could summarize our Solar System's development by saying that Choose one:
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D. It formed solely from atoms generated by the Big Bang.

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From the history of the earth and it's solar system, options A to C are all correct ways of summarizing the Solar System's development. Because they have been proven to be in sync with reality of the solar system whereas the Bing bang theory has not done so as we know it says the universe was birthed from a huge explosion called big bang.

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Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
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Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

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2 years ago
A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th
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Answer:

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Explanation:

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Enthalpy of vaporization =  37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

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