The answer would be A, 100g since it’s equal
<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.
Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.
Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)
Answer:
Lavoisier; Newlands; Moseley
Explanation:
In 1789, Antoine Lavoisier grouped the elements into gases, nonmetals, metals, and earths.
In 1865, John Newlands developed the Law of Octaves. He stated that "any given element will exhibit analogous behaviour to the eighth element following it in the table."
In 1914, Henry Moseley found a correlation between the X-ray wavelength of an element and its atomic number. He was then able to restructure the periodic table according to atomic numbers.
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
