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Leno4ka [110]
3 years ago
6

Calculate the heat energy released when 21.1 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.

Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
3 0
Heat capacity yea urjjrjrjrbdnsnnend
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A mole of Big Macs stacked would reach from the earth to the moon and back how many times?
OLga [1]
A mol=6.02*10^23
so
distance from earth to moon is 230,100 miles
1 big mac is about 2 inches
1 mile=5280 feet=63360inches

230,100 times 63360=14,579,136,000 inches is distance
6.02 times 10^23 times 2=x times 14,579,136,000
divide both sides by 2
6.02 times 10^23=x times 7,289,568,000
divide both sides by 6.02
10^23=x times 1210891694.35159
divide both sides by  1210891694.35159
825583769024485.8
about 825,583,769,024,486 times

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3 years ago
Circle the letter of cach sentence that is true about silica.
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Common additives to drinking water include elemental chlorine, chloride ions, and phosphate ions. Recently, reports of elevated
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3 years ago
2.When copper (II) nitrate reacts with sodium hydroxide, copper (11) hydroxide is produced.
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2Cu + S ~~~> Cu2S Copper (C) reacts with sulfur (S) to form copper sulfide as shown in the equation. A scientist adds 12.7 grams of Cu to 3.2 grams of S to start the reaction.

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2 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

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.

5 0
2 years ago
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