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Leni [432]
2 years ago
7

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 li

ter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4. CO (g) 3 H2 (g) rightwards harpoon over leftwards harpoon CH4 (g) H2O (g)
Chemistry
1 answer:
Temka [501]2 years ago
7 0

Answer:

can you be more clear with your question :

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FromTheMoon [43]

Answer:

Inner core - A

Outer core - B

Mantle - C

Asthenosphere - F

Lithosphere - E

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7 0
1 year ago
The conversion of 37 ˚C to K
77julia77 [94]

Answer:

310.15

Explanation:

c + 273.15 = k

37+ 273.15 = 310.15

8 0
3 years ago
Read 2 more answers
What mass of radon is in 8.17 moles of radon?
Len [333]

Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon  = 8.17moles

Unknown:

Mass of radon  = ?

Solution:

To solve this problem, we use the expression below:

      Number of moles = \frac{mass}{molar mass}  

Molar mass of radon  = 222g/mol

Now insert the parameters and solve;

    Mass of radon  = Number of moles x molar mass

                              = 8.17 x 222

                              = 1813.74g

4 0
3 years ago
PLEASE HELP DUE SOON> WILL MARK BRAINLIEST 50 POINTS
Rainbow [258]
Methane CH4
ethane C2H6
propane C3H8
Butane C4H10
pentane C5H12
8 0
3 years ago
Read 2 more answers
Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this
maw [93]

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol

Thus the enthalpy change for the given reaction is -705.55kJ

7 0
3 years ago
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