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kogti [31]
2 years ago
14

Consider the reaction 3O2(g) 2O3(g) At 298 K, the equilibrium concentration of O2 is 1.6 x 10^-2 M, What is the equilibrium cons

tant of the reaction at this temperature?
Chemistry
1 answer:
Sedaia [141]2 years ago
8 0

Answer:

Kc = 105062.5 at 298K

Explanation:

First of all, we state the equilibrium:

3O₂(g) ⇄  2O₃(g)

We know data about equilibrium concentration of oxygen. We suppose 1 mol of oxygen at the begining. During the reaction, x moles have reacted.

As ratio is 2:3, we can determine how many moles of ozone have been produced.

(x . 2)/3

So we have the final concentration of oxygen, so, let's find out x

1 - x = 0.016 moles

x = 0.984 moles

Then, the [O₃] in equilibrium will be (0.984 . 2) /3 = 0.656.

We supose a volume of 1 L, so we have molar concentration to determine Kc. Let's state the expression for it:

Kc = [O₃]² / [O₂]³

Kc = 0.656² / 0.016³ → 105062.5

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zysi [14]
The molar mass of the compound potassium nitrate, KNO3 is equal to 101.1032 g/mol. Then, we determine the number of moles present in the given amount,
                       n = 11.75g / (101.1032 g/mol) = 0.116 mol
Then, molarity is calculated by dividing the number of moles by the volume of the solution. The answer is therefore 0.058 M. 
3 0
3 years ago
How many sulfur atoms are generated when 9.42 moles of H2S react according to the following equation: 2H2S+SO2→3S+2H2O
8_murik_8 [283]

Answer:

A) 8.51 × 10²⁴  

Explanation:

1. Gather all the information

            2H₂S + SO₂ ⟶ 3S + 2H₂O

n/mol:   9.42

2. Calculate the moles of S atoms

The molar ratio is 3 mol S:2 mol H₂S

\text{Moles of S} = \text{9.42 mol H$_{2}$S} \times \dfrac{\text{3 mol S }}{\text{2 mol H$_{2}$S }} = \text{14.13 mol S}

3. Calculate the atoms of S

\text{Atoms of S } = \text{14.13 mol S} \times \dfrac{6.022 \times 10^{23}\text{ S atoms}}{\text{1 mol S}} = \mathbf{8.51 \times 10^{24}}\textbf{ S atoms}

 

6 0
3 years ago
Read 2 more answers
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
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When the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solu
gtnhenbr [62]

Answer:

The unknown solution had the higher concentration.

Explanation:

When two solutions are separated by a semi-permeable membrane, depending on the concentration gradient between the two solutions, there is a tendency for water molecules to move across the semi-permeable in order to establish an equilibrium concentration between the two solutions. This movement of water molecules across a semi-permeable membrane in response to a concentration gradient is known as osmosis. In osmosis, water molecules moves from a region of lower solute concentration or higher water molecules concentration to a region of higher solute concentration or lower water molecules concentration until equilibrium concentration is attained.

Based on the observation that when the glucose solution described in part A is connected to an unknown solution via a semipermeable membrane, the unknown solution level rises, it means that water molecules have passed from the glucose solution through the semipermeable membrane into the unknown solution. Therefore, the solution has a higher solute concentration than the glucose solution.

3 0
3 years ago
Give 3 examples of solutions, suspensions and colloids..
Vlad [161]
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  • <u>examples of colloids are </u><u>whipped cream, mayonnaise, milk, </u>
5 0
2 years ago
Read 2 more answers
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