Answer:
3.51 g of oxygen per gram of gasoline is required.
Explanation:
Solution:
First of all we will write the balance chemical equation.
C8H18 + 12.5O2 → 8CO2 + 9H2O
This equation shows that,
1 mole of gasoline react with 12.5 mole of oxygen for complete burning.
mass of one mole of gasoline = 8×12 + 18×1 = 114 g
mass of 12.5 mole of oxygen = 12.5 (16×2) = 400 g
Formula:
mass of oxygen per gram of gasoline = (400 / 114) = 3.51
so, 3.51 g of oxygen require for per gram of gasoline.
Answer:
Saturn. Saturn is the sixth planet from the Sun and the second-largest in the Solar System, after Jupiter. It is a gas giant with an average radius of about nine times that of Earth. It only has one-eighth the average density of Earth; however, with its larger volume, Saturn is over 95 times more massive.
Answer:
5.158 × 10²³ atoms K
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
33.49 g K
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of K - 39.10 g/mol
<u>Step 3: Convert</u>
<u />
= 5.15797 × 10²³ atoms K
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig figs and round.</em>
5.15797 × 10²³ atoms K ≈ 5.158 × 10²³ atoms K
