<h2>1.15 moles</h2>
Explanation:
The ideal gas law is written as 
where 
are the pressure, volume and temperature,
is the number of moles of gas; and 
is the ideal gas constant. It is the same for all gases.
Given,
Since,
So,the number of moles is 
Answer:
0.52 g of KNO₃ are contained in 19.7 mL of diluted solution.
Explanation:
We can work on this problem in Molarity cause it is more easy.
Molarity (mol/L) → moles of solute in 1L of solution.
100 mL of solution = 0.1 L
We determine moles of solute: 44.7 g . 1mol /101.1 g = 0.442 mol of KNO₃
Our main solution is 0.442 mol /0.1L = 4.42 M
We dilute: 4.42 M . (11.9mL / 200mL) = 0.263 M
That's concentration for the diluted solution.
M can be also read as mmol/mmL, so let's find out the mmoles
0.263 M . 19.7mL = 5.18 mmol
We convert the mmol to mg → 5.18 mmol . 101.1 mg / mmol = 523.7 mg
Let's convert mg to g → 523.7 mg . 1 g / 1000 mg = 0.52 g
We want:
S(s) + O2(g) --> SO2(g)
So the following are the given:
1) S(s) + 3/2O2(g) --> SO3(g) ∆H = -395.8 kJ/mole
2) 2SO2 + O2 --> 2SO3(g) ∆H = -198.2 kJ/mole
Reverse Equation 2) and then divide by 2
SO3(g) --> SO2(s) + 1/2O2(g) ∆H = +99.1 kJ/mole
Add Equation 1)
S(s) + O2(g) --> SO2(g) ∆H = -296.7 kJ/mole
Answer:
1.21 mol KClO₃
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Mole Ratio
<u>Stoichiometry</u>
- Analyzing reactions rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[rxn] 2KClO₃ → 2KCl + 3O₂
[Given] 58.3 g O₂
[Solve] mol KClO₃
<u>Step 2: Identify Conversions</u>
[rxn] 2 mol KClO₃ → 3 mol O₂
[PT] Molar Mass of O: 16.00 g/mol
Molar Mass of O₂: 2(16.00) = 32.00 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.21458 mol KClO₃ ≈ 1.21 mol KClO₃
It should just be one for it to be balanced.
