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Sav [38]
3 years ago
6

Consider the reversible reaction A ( g ) − ⇀ ↽ − B ( g ) Which K values would indicate that there is more B than A at equilibriu

m? K = 9 × 10 – 5 K = 0.4 K = 5000 K = 8 × 10 7
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
4 0

Answer:

5000 and 8\times 10^{7} indicate that there is more B than A at equilibrium

Explanation:

For the given reaction: K=\frac{[B]}{[A]}

where [B] and [A] represents equilibrium concentration B and A respectively. K represents equilibrium constant

More B than A at equilibrium means, [B] > [A]

So, K=\frac{[B]}{[A]}>1

As, both 5000 and 8\times 10^{7} are greater than 1 therefore these two K values indicate that there is more B than A at equilibrium

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marishachu [46]

Answer:

C) 1 x 10-10 M

Explanation:

To solve this question we must use the equation:

Kw = [H+] [OH-]

<em>Where Kw is the equilibrium dissociation of water = 1x10-14</em>

<em>[H+] is the molar concentration of hydronium ion = 1x10-4M</em>

<em>[OH-] is the molar concentration of hydroxyl ion</em>

<em />

Replacing:

1x10-14= 1x10-4 [OH-]

<em>[OH-] = 1x10-14 / 1x10-4M</em>

<em>[OH-] = 1x10-10 M</em>

Right option is:

<h3>C) 1 x 10-10 M </h3>

7 0
3 years ago
Indigenous rocks form by the cooling and hardening of melted rock.* True False​
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Answer: True

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3 0
3 years ago
How do ions form bonds, and describe the structure of the resulting compound.​
Artemon [7]

Answer:

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6 0
2 years ago
A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and
AleksandrR [38]

Answer:

a)Pheptane = 24.3 torr          

Poctane = 5.12 torr    

b)Ptotal vapor = 29.42 torr

c)  81 % heptane

    19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

                               Xa = mole fraction of A

                               Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial  vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor  in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

          Xa = Pa/Pt where Xa = mol fraction of  in the vapor

                                       Pa = partial pressure of A as calculated above

                                        Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)    

a)                  

 MW heptane = 100.21 g/mol

 MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

                             Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part   we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane  and then we can calculate the masses:

0.82 mol x 100.21  g/mol = 82.2 g

0.17 mol x 114.23 g/mol =  19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol)  + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then  having a mol fraction of 0.83  means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x  114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8  vs 10.9 torr).

4 0
3 years ago
How many moles of koh are contained in 250 ml of a 2.0 m solution of koh
kotykmax [81]
.5 mols
Assuming that your 2.0 m is an M for molarity
I used the formula M=number of mold/L
Converted 250mL to .250L by dividing by 1000
4 0
3 years ago
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