Answer:
the concentration of PCl5 in the equilibrium mixture = 296.20M
Explanation:
The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.
The detaied steps is as shown in the attached file.
Answer:
An equal force acts in the opposite direction.
Explanation:
Explanation:
For Mg, (1.2 g Mg/24 g Mg) = 0.05 mol Mg.
For Cl, (3.55 g Cl/35.5 g Cl) = 0.1 mol Cl
So the ratio now is
Mg:Cl = 0.05 : 0.1 = 1:2
I got the 1:2 ratio by dividing both by the smallest number, which is 0.05 mol. Therefore, the empirical for formula of the substance is 
Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is not complete and lacks necessary data to solve. However, I have found a similar question and I will be using its data to solve this question for the sake of understanding and concept.
Solution:
Equilibrium Reaction:
CaO(s) + H2O(g) -->Ca(OH)2(s)
We need to find the reaction quotient for this question:
Q =
Here, only the pressure of the gaseous reactant will be used and here H20 is the only reactant which is gaseous.
And we are given that, vapor pressure of water is = 0.106 mmHg
So,
Now, we need to convert it into atm
so, 1atm = 760 mmHg
0.106 mmHg = 0.106/760 atm
0.106 mmHg = 1.394 x
atm
Plugging in the values in the equation, we get:
Q =
Q = 
Q = 7173.60 
<h3>
Answer:</h3>
23.459 g NaNO₂
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄
[Given] 24.14714 g Na₂SO₄
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂
Molar Mass of Na - 22.99 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol
Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We need 5 sig figs (instructed).</em>
23.4587 g NaNO₂ ≈ 23.459 g NaNO₂