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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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Particles of matter how have both potential and kinetic energy is that true or false?
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4 years ago
What is the molality of a solution made by dissolving 15.20 g of i2 in 1.33 mol of diethyl ether, (ch3ch2)2o?
Paraphin [41]
The  molarity   of solution  made  by  dissolving  15.20g  of i2  in 1.33 mol  of diethyl ether (CH3CH2)2O  is    =0.6M

   calculation

molarity  =moles of solute/  Kg of the  solvent

mole  of the solute  (i2)  =  mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore=  15.2 g/253.8 g/mol  =  0.06  moles


calculate the Kg of solvent  (CH3CH2)2O
mass =  moles  x  molar mass
molar mass  of  (CH3CH2)2O= 74 g/mol

mass  is therefore = 1.33 moles  x  74 g/mol =  98.42 grams
in Kg = 98.42 /1000 =0.09842  Kg

molarity  is therefore = 0.06/0.09842 = 0.6 M

3 0
4 years ago
Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:
Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

               2 ICl(g)   ⇄   I₂(g)    +    Cl₂(g)

Initially      0.20              -               -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React          x              x/2               x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

           (0.20 - x)          x/2             x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0
3 years ago
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