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Anastaziya [24]
3 years ago
13

A system was prepared with NH3 = O2 = 3.60 M as the only components initially. At equilibrium N2O4 is 0.60M. Calculate the value

of the equilibrium constant Kc for the reaction
Chemistry
1 answer:
frozen [14]3 years ago
8 0

Answer:

Kc = 0.0216

Explanation:

To do this, we need to write the equilibrium reaction which is the following:

NH3 + O2 <------> N2O4 + H2O

And now, let's balance the equation

4NH3 + 7O2 <--------> 2N2O4 + 6H2O

The general expression for Kc is:

Kc = [N2O4]²[H2O]⁶ / [O2]⁷[NH3]⁴

Now, in order to know the concentrations, we need to do an ICE chart here, so we can know the final concentrations of all species here:

         4NH3 + 7O2 <--------> 2N2O4 + 6H2O

I:          3.6        3.6                      0            0

C:         -4x       -7x                      +2x        +6x

E:      3.6 - 4x   3.6 - 7x              0.6          +6x

According to this, we know that N2O4 is 0.6M, but in the change we have that it gained 2x so the value of x will be:

2x = 0.6

x = 0.6/2 = 0.3

Therefore, the value of x is 0.3 ans we can know now the concentrations of all species in equilibrium, and then, the kc:

[NH3] = 3.6 - 4(0.3) = 2.4 M

[O2] = 3.6 - 7(0.3) = 1.5 M

[H2O] = 6(0.3) = 1.8 M

Now the value of Kc would be:

Kc = (0.6)² (1.8)⁶ / (2.4)⁴ (1.5)⁷

Kc = 0.0216

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Answer: 11.0 g of calcium will react with 10.0 grams of water.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2O

\text{Number of moles}=\frac{10.0g}{18g/mol}=0.55moles

The balanced chemical equation is:

Ca+2H_2O\rightarrow Ca(OH)_2+H_2

According to stoichiometry :

2 moles of H_2O require = 1 mole of Ca

Thus 0.55 moles of H_2O require=\frac{1}{2}\times 0.55=0.275moles  of Ca  

Mass of Ca=moles\times {\text {Molar mass}}=0.275moles\times 40g/mol=11g

Thus 11.0 g of calcium will react with 10.0 grams of water.

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3 years ago
The indicator methyl red has different molecular structures at high and low pH.
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As is used in titrations of acid-base reactions the indicator change in colour. Is red when the pH < 4.4 (Acidic Solutions) and is yellow when pH > 6.2 (Neutral-Basic solutions).

A change in colour means the structure of the indicator is changing with pH. Thus, the answer is:

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How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant
nika2105 [10]
<h3>Answer:</h3>

0.0253 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

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

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 0.025305 \ mol \ H_2O

<u>Step 4: Check</u>

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

0.025305 mol H₂O ≈ 0.0253 mol H₂O

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