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Jet001 [13]
3 years ago
10

3. Sulfur dioxide (SO2) can be reduced by hydrogen disulfide (H2S), producing sulfur in its elemental form and water (H2O). Usin

g Hess’s law and the reactions below, determine the overall enthalpy for the reaction between SO2 and H2S. Make to clearly show your work.
H2 (g) + S (s) → H2S (g) ΔH = –20.6 kJ
SO2 (g) → S (s) + O2 (g) ΔH = +296.8 kJ
2H2 (g) + O2 (g) → 2H2O (g) ΔH = –285.8 kJ
Chemistry
1 answer:
Dafna1 [17]3 years ago
8 0
The reaction between h2s and so2 results to h20 and elemental sulfur. The final reaction is SO2 +2H2S = 3S+ 2H2O. Using Hess law, the first equation is reversed and multiplied by 2, the second and the third equation remains the same. Calculating the overall enthalpy, the answer is 52.2 kJ. 
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Tough and malleable form of iron is
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3 years ago
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Need help setting the problem up
Sveta_85 [38]

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

5 0
3 years ago
A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
VMariaS [17]

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

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(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g

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3 years ago
Describe the three types of system: a.open: b.closed: c. Isolated:
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An open system is when there is a transfer of energy and matter with its surroundings.

A closed system is where there is a transfer of energy, may it be heat or work, but not matter with its surroundings.

An isolated system is where there is no transfer of energy or matter with its surroundings.

4 0
3 years ago
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