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Sloan [31]
3 years ago
7

How many molecules of Oxygen are consumed if 24 molecules of Carbon Dioxide are produced?

Chemistry
1 answer:
Nat2105 [25]3 years ago
6 0

Answer:

Explanation:

C + O2 → CO2

Mole of C = 24 g/(12 g/mole)

Mole of C = 2 mole

Mole of molecular O2 = 74 g/(32 g/mole)

Mole of molecular O2 = 2.3125 mole

Since mole of C < mole of O2, then C being the limiting reagent.

From the reaction, it shows that mole ratio between C and O2 = 1 : 1.

So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.

Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.

Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.

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Answer: Equilibrium concentration of [Cl^-] at 50^0C is 4.538 M

Explanation:

Initial concentration of CoCl_2 = 0.056 M

Initial concentration of Cl^- = 4.60 M

The given balanced equilibrium reaction is,

               COCl_2+2Cl^-\rightleftharpoons [CoCl_4]^{2-}+6H_2O

Initial conc.          0.056 M      4.60 M        0 M       0 M

At eqm. conc.     (0.056-x) M   (4.60-2x) M   (x) M    (6x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CoCl_4]^{2-}\times [H_2O]^6}{[CoCl_2]^2\times [Cl^-]^2}

Given : equilibrium concentration of [CoCl_4]^{2-} =x =  0.031 M

Concentration of Cl^- = (4.60-2x) M  = (4.60-2\times 0.031) =4.538 M  

Thus equilibrium concentration of [Cl^-] at 50^0C is 4.538 M

8 0
4 years ago
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solniwko [45]

Answer:

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Explanation:

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Na+ Br2 = NaBr

Na × 2 = Na2

Na × 2 = Na2

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2Na + Br2 = 2NaBr

Hope this helps.

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