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alexandr402 [8]
3 years ago
11

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0100 M O2. At this temperature, Kc

equals 1.4 × 102 for the reaction: 2 CO(g) + O2(g)⇌2 CO2(g). What is the equilibrium concentration of CO?
A) 7.1 × 10-7 M.
B) 8.4 × 10-4 M.
C) 1.4 × 10-2 M.
D) 1.2 × 10-1 M.
Chemistry
2 answers:
Naya [18.7K]3 years ago
8 0
The correct answer to this question is letter "B) 8.4 × 10-4 M."

2CO+O2<->2CO2

-x       -x        +x

It started as 2CO2 so +x, then it decomposed into 2CO and O2, so-x

2CO-oxidized, O2-Oxidized, 2CO2-reduced


the easiest way is to look at the equation, so if we have some reactant and no product 

A<->2C+D

-x    +2x +x
so if we had product concentration, and no reactant concentration 3A<-> C+D +3x -x-x <span>
</span>
salantis [7]3 years ago
5 0

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

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

The given chemical reaction follows:

2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g)

The expression of K_{c for above reaction is:

K_c=\frac{[CO_2]^2}{[CO]^2[O_2]}

We are given:

K_c=1.4\times 10^2

[CO_2]=0.0010M

[O_2]=0.0100M

Putting values in above equation, we get:

1.4\times 10^2=\frac{(0.0010)^2}{[CO]^2\times (0.0100)}

[CO]=\sqrt{\frac{(0.0010)^2}{0.0100\times 1.4\times 10^2}}

[CO]=8.4\times 10^{-4}M

Hence, the correct answer is Option B.

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An alkene X undergoes ozonolysis and gives two compounds Y and Z of molecular formula CaHO. Y and Z are functional isomers of ea
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i. The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.

ii. The structural formula of Y is CH_3-CO-CH_3 and Z is CH_3-CH_2-CHO.

iii. Alkenes, upon catalytic hydrogenation, form alkanes. This will occur in the presence of Nickel as the catalyst.

iv. The process of ozonolysis is useful in the field of pharmaceutics.

v. The test of unsaturation can be performed by passing a compound through Bromine water.

<h3>What is ozonolysis?</h3>

Ozonolysis is a reaction used in organic chemistry to determine the position of a carbon-carbon double bond in unsaturated compounds.

i. The given alkene X, that is subject to ozonolysis would be 2-methyl-2-pentene. Upon exposure to ozone, an ozonide is initially formed, after which it is broken down into 2 products - acetone and propanal, both with the molecular formula C₃H₆O.

The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.

ii. The formulas of Y is CH_3-CO-CH_3 and Z is CH_3-CH_2-CHO. They are functional isomers as they have the same molecular formula but different functional groups - ketone and aldehyde.

iii. When alkenes undergo catalytic hydrogenation, they form alkanes, X will form 2 methyl petane on reaction with hydrogen gas in presence of Ni.

iv. The ozonolysis is used for the industrial-scale synthesis of pharmaceuticals.

v. The unsaturation of compound X can be proved by the bromine water test. As on reaction with it, the brown colour of bromine water becomes colourless due to the formation of dibromo alkane.

Learn more about the ozonolysis here:

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5 0
2 years ago
If a sample of magnesium with a mass of 35.0 grams reacts with 35.0 grams of oxygen how much magnesium oxide will be produced?​
slamgirl [31]

Answer:

58.0 g of MgO

Explanation:

in a perfect world, 70 g, however we don't live in a perfect world

The equation of reaction

2Mg + O₂ --> 2MgO

first find which element is limiting:

35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO

35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO

This means Mg is the limiting factor, so you will be using this moles to find grams of MgO

1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO

8 0
3 years ago
The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc
jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br_2(g

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

+\frac{1d[Br_2]}{dt}=0.151

Thus the rate of appearance of Br_2 is 0.151

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