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<h3>
Answer:</h3>
3.38 × 10²⁴ molecules CO₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 5.61 moles CO₂
[Solve] molecules CO₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</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>
3.37834 × 10²⁴ molecules CO₂ ≈ 3.38 × 10²⁴ molecules CO₂
Answer:
C. Lose three electrons to have a full outer shell
Explanation:
Al is in Group 13 of the Periodic Table, so it has three valence electrons.
It must either lose three electrons or gain five to achieve a stable octet.
It is easier to lose three electrons than it is to gain five, so Al loses three electrons.
D. is wrong, for the same reason.
A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.
C. is wrong. Al is a metal, so it will lose electrons in a reaction.
Answer:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
Answer:
Helium
Explanation:
The first ionization energy varies in a predictable way across the periodic table.