<h3>
Answer:</h3>
0.34 mol S
<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>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
11 g S
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of S - 32.07 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
0.343 mol S ≈ 0.34 mol S
Answer:
Rb: [Kr] 5s
Step-by-step explanation:
Rb is element 37, the first element in Period 5.
It has one valence electron, so its valence electron configuration is 5s.
The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.
The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.
Answer:
Cool air moves down while hot air rises
Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:
(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
