Does water gain or release energy when evaporated into water vapor?

Chromium (III) oxide reacts with hydrogen sulfide gas to form chromium (III) sulfide and water. To produce 421 g of cr2s3, how m

maw [93]

Cr₂O₃ ( s ) + 3H₂S ( g ) → Cr₂S₃ ( s ) + 3H₂O ( l )

mol Cr₂S₃ = 421 : 200.19 g/mol = 2.103

mol Cr₂O₃ ≈ mol Cr₂S₃ = 2.103 ( equivalent coefficient)

mass Cr₂O₃ = 2.103 x 151.99 g/mol = 319.63 gr

6 0

1 year ago

Which element are cations? how do you know?

Paladinen [302]

If it has a positive charge it is a cation if negative it is an anion .

I attached a chart that will help you know the charges of the elements

8 0

2 years ago

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HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

2 years ago

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A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem

Ludmilka [50]

Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L

⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K

⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

8 0

2 years ago

Mwehehe poide bang pakiss?

Contact [7]

Answer:

what confused

Explanation:

6 0

2 years ago

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