Answer:
the empirical (lowest raios) is
C2H4Cl
Explanation:
A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.
What is the empirical formula of this compound?
the compound has ONLY C, H, and Cl
the % Cl = 100% - 24.27% -4.03% = 71.7%
in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H
the number of moles are Cl=71.7/70.91 =1.01= ~ 1
C = 24.27/12.0 = 2.02 =~ 2
H = 403/1.01 = 3.97 =~ 4
so the empirical (lowest raios) is
C2H4Cl
Answer:
Explanation:
We'll assume there is an excess of silver nitrate, so that all 12.0 moles of the magnesium (Mg) will react.
The balanced equation tells us we'll obtain 2 moles of Ag for every 1 mole of magnesium, for a molar ratio of 2/1.
Starting with 12.00 moles Mg, we would therefore hope to find twice that, or 24.00 moles of Ag.
To convert to grams, find the molar mass of Ag from the periodic table.
Ag has a molar mass of 107.9 (to 4 sig figs) grams/mole.
(24.00 moles)*(107.9 grams/mole) = 2590 grams (4 sig figs)
Hands off, it's mine.
Dmitri Mendeleev
<span> Dmitri Mendeleev and Lothar Meyer individually came up with their own periodic law "when the elements are arranged in order of increasing atomic mass,
certain sets of properties recur periodically.</span>
<h3>
Answer:</h3>
<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>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂
[Given] 20 mol Al₂O₃
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Al₂O₃ → 4 mol Al
<u>Step 3: Stoich</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 1 sig fig.</em>
Since our final answer already has 1 sig fig, there is no need to round.
I: Current
V: Voltage
R: resistance
you’re welcome ;)
