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Sliva [168]
3 years ago
12

Please help I will mark Brainly!

Chemistry
1 answer:
olga nikolaevna [1]3 years ago
8 0
I think C. Ionic compounds are very difficult to break apart because they are fully negative or positive charges, like magnets.
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A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water
Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

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= 6 : 1 : 12

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molecular weight = n ( 6 x 12 + 12x 1 + 16)

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Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)
Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g)              K₁= ?                       (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g)                 K₂= 3.61                 (2)

2 A(g) + D(g)  ⇄ C(g)             K₃= 7.19                 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3)  algebraically  to get our desired equilibrium (1).

We are allowed to reverse  reactions, in that case we take the reciprocal of K as our new K' ; we can also  add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this  in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3)  and we want D as a reactant . That  suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B  as product in our  equilibrium (1) . Finally we would add (2) and (3) to get  (1) which is our final  goal.

2C(g)             ⇄  2A(g) + 2B(g)  K₂´= ( 1/ 3.61 )²  

                                   ₊

2 A(g) + D(g)  ⇄     C(g)               K₃ = 7.19  

<u>                                                                                    </u>

C(g) + D(g)     ⇄    2B(g)       K₁ = ( 1/ 3.61 )²   x  7.19

                                             K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves  gases.

To compute the last part lets setup the following mnemonic  ICE table to determine the quantities at equilibrium:

pressure (atm)        C             D           B

initial                     1.64          1.64         0

change                    -x             -x        +2x

equilibrium          1.64-x         1.64-       2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)²  where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x  we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0
3 years ago
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