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dimulka [17.4K]
1 year ago
8

Hydrogen gas and bromine gas react to form hydrogen bromide gas. How much heat (kJ) is released when 155 grams of HBr is formed

in this reaction? ΔH° = -72 kJ.
Chemistry
1 answer:
VashaNatasha [74]1 year ago
8 0

Answer:

-37.63KJ

Explanation:

First, we write an equation of reaction:

H + Br ——-> HBr

Now we are told that 155g of HBr is formed. We can calculate the number of moles of HBr formed as follows:

We simply divide the mass of the HBr formed by the molar mass of HBr. The molar mass of HBr is 81g/mol

The number of moles of HBr formed is thus 81/155 = 0.52 moles

The amount of heat released is thus 0.52 * (-72) = -37.63KJ

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0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

8 0
2 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
1 year ago
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r
Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0
1 year ago
What two-step process separated the cans into aluminum, steel, and tin?
Vikentia [17]

Answer:

Magnet

Durability and heaviness.(texture)

Explanation:

Magnet can be use to separate Aluminum from mixture of steel and aluminum.

Though aluminum and steel look alike but magnet can be use to separate it.

If the can attract the magnet or magnet stick to the can, it is a steel can. Aluminum does not stick to magnet.

A mixture of Aluminum and tin can also be separated by magnet.

Tin attract magnet but tin is more durable, heavy and does not corrode easily.

When u touch the three cans, tin is heavy and durable.

4 0
2 years ago
A sample of 22K gold contains the following: 22 g gold, 1.0 g silver, and 1.0 g copper. What is the percent gold in the sample?
iris [78.8K]

Answer:

35 percent

Explanation:

8 0
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