Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
<u>HI :)</u>
True
<em>Hope</em><em> this</em><em> helps</em><em> </em><em>:</em><em>)</em>
<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:
Either Carnivores or Heterotrophic.
Explanation:
Carnivorous organisms are the more obvious answers in this, but if you want a bit of pizazz, throw in Heterotrophic Fungi.
Answer:a) 2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction
b) 8 (4 moles CO2 per mole butane)
Explanation:
could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O