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Novosadov [1.4K]
3 years ago
10

I need help! In an equilibrium reaction with a Keq of 1 × 108 A. products are favored. B. the reaction is nonspontaneous. C. the

reaction is endothermic. D. reactants are favored.
Chemistry
2 answers:
scZoUnD [109]3 years ago
7 0
<span>When K is > than 1 products are favored. When q is < less than 1 reactants are favored. 1 x 108 = 108 which is > 1 so products are favored.</span>
Amiraneli [1.4K]3 years ago
3 0

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants. It is represented as K_{eq}

For the general reaction:

A+B\rightarrow C+D

Expression for equilibrium constant is given as:

K_{eq}=\frac{[C][D]}{[A][B]}

Conditions for K_{eq} are:

K_{eq}>1, then products are favored

K_{eq}=1, then forward reaction is equal to backward reaction

K_{eq}, then reactants are favored

As, K_{eq}=1\times 10^8 and is very much greater than 1, so products will be favored.

Hence, the correct answer is Option A.

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nataly862011 [7]

Answer:

oxygen

Explanation:

because the 2nd shell is not complete which is suppose to be 8 and since oxygen is 8 it first shell is 2 which is complete and the second shell which is 6 is not complete because we all know that 2+6=8 but the standard shell is

K-2

L-8

M-8

3 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
Calculate the percentage by mass of oxygen in trioxonitrate (v) acid, (HNO3). [ H=1, N=14, O=16]​
Vadim26 [7]

Answer:

Percentage by mass of oxygen = 76.20% (Approx)

Explanation:

Given:

HNO3

H=1, N=14, O=16]

Find:

Percentage by mass of oxygen

Computation:

HNO3

Total mass = 1 + 14 + 3(16)

Total mass = 63

Mass of oxygen = (3)(16) = 48

Percentage by mass of oxygen = [48/63]100

Percentage by mass of oxygen = 76.20% (Approx)

4 0
3 years ago
A solution has a pH of 12. This solution is:
kolezko [41]

Answer:

basic

Explanation:

pH 7: neutral

12>7

so it is basic

(if <7 than acidic)

3 0
3 years ago
Which common material is an example of a polyamide?
Vlad1618 [11]

Nylon 6,6 is a common example of a polyamide.

<em>Polyamides</em> are polymers that contain <em>repeating amide (-CO-NH-) linkages</em>.

The structure of Nylon 6,6 is  

[-NH-(CH_2)_6-<u>NH-CO</u>-(CH_2)_4-CO-]_<em>n</em>

where <em>n</em> is a large number.

The numbers in the name showow that there are six carbon atoms on either side of an amide linkage.

5 0
3 years ago
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