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vazorg [7]
2 years ago
13

The equilibrium constant Kc for the reaction

Chemistry
1 answer:
PIT_PIT [208]2 years ago
3 0

Answer:

[N2] = 0.3633M

[H2] = 1.090M

[NH3] = 0.2734M

Explanation:

Based on the reaction of the problem, Kc is defined as:

Kc = 0.159 = [NH3]² / [N2] [H2]³

<em>Where [] are the equilibrium concentrations.</em>

The initial concentrations of the reactants is:

N2 = 1.00mol / 2.00L = 0.500M

H2 = 3.00mol / 2.00L = 1.50M

When the equilibrium is reached, the concentrations are:

[N2] = 0.500M - X

[H2] = 1.50M - 3X

[NH3] = 2X

<em>Where X is reaction quotient</em>

Replacing in the Kc equation:

0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³

0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴

0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²

0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0

Solving for X:

X = 0.1367. Right solution.

X = 1.8286. False solution. Produce negative concentrations

Replacing:

[N2] = 0.500M - 0.1367M

[H2] = 1.50M - 3*0.1367M

[NH3] = 2*0.1367M

The equilibrium concentrations are:

<h3>[N2] = 0.3633M</h3><h3>[H2] = 1.090M</h3><h3>[NH3] = 0.2734M</h3>

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What is the equilibrium expression for the reatcion bellow<br><br> 2SO3(g)&lt;=&gt;O2(g)+2SO2(g)
Luda [366]

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

8 0
3 years ago
If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?
Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

412.072 g Cl₂ ≈ 412 g Cl₂

5 0
3 years ago
there about 5000 elements, which combine to form the vast number of diffrent substances in the world around us
marissa [1.9K]

Answer: false 100

Explanation:

if this is the true or false test here

5 0
3 years ago
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
meriva

Answer:

50 g Sucrose

Explanation:

Step 1: Given data

  • Mass of solution: 2000 g
  • Concentration of the solution: 2.5%

Step 2: Calculate the mass of sucrose needed to prepare the solution

The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:

2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose

5 0
3 years ago
Which statement supports one of Dalton's contributions to the atomic theory?
Mnenie [13.5K]
<span>C) The element nitrogen is made up of carbon and nitrogen</span>
6 0
3 years ago
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