Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
There are 1000 grams in a kg.
To convert g to kg, dovide by 1000.
3.5/1000= 0.0035 kg
Final answer: D
Answer : The final temperature of the solution in the calorimeter is, 
Explanation :
First we have to calculate the heat produced.

where,
= enthalpy change = -44.5 kJ/mol
q = heat released = ?
m = mass of
= 1.52 g
Molar mass of
= 40 g/mol

Now put all the given values in the above formula, we get:


Now we have to calculate the final temperature of solution in the calorimeter.

where,
q = heat produced = 1.691 kJ = 1691 J
m = mass of solution = 1.52 + 35.5 = 37.02 g
c = specific heat capacity of water = 
= initial temperature = 
= final temperature = ?
Now put all the given values in the above formula, we get:


Thus, the final temperature of the solution in the calorimeter is, 
<h3>
Answer:</h3>
10.6 mol NO
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O
[Given] 13.2 mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 5 mol O₂ → 4 mol NO
<u>Step 3: Stoich</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
10.56 mol NO ≈ 10.6 mol NO
Answer:
Explanation:
Total moles = 300 + 500 = 800 moles
mole fraction of carbon dioxide = 300 / 800 = 3 / 8
water = 500 / 800 = 5 / 8
Total pressure = 21 atm
partial pressure of water = total pressure x mole fraction of water
= 21 x 5 / 8
= 13.125 atm