Answer is: A) 124 s.
c₀ = 3 mol/L.
c₁ = 0,700 mol/L.
k = 8,8·10⁻³ 1/M·s.
Integrated second order rate law is: 1/c₁ = 1/c₀ + k·t.
k·t = 1/0,700 - 1/3.
0,0088·t = 1,095.
t = 1,095 ÷ 0,0088.
t = 124 s.
c₀ - <span>initial concentration.
c</span>₁ - <span> concentration at a particular time.
k - </span><span>the rate constant.
t - time.</span>
Answer:
Empirical formula: CH₃O
Empirical formula mass = 31 g/mol
Explanation:
Data Given:
Molecular Formula = C₁₀H₃₀O₁₀
Empirical Formula = ?
Empirical Formula mass =
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule
As
C₁₀H₃₀O₁₀ Consist of 10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.
Now
Look at the ratio of these three atoms in the compound
C : H : O
10 : 30 : 10
Divide the ratio by two to get simplest ratio
C : H : O
10/10 : 30/10 : 10/10
1 : 3 : 1
So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1
So the empirical formula will be
Empirical formula of C₁₀H₃₀O₁₀ = CH₃O
Now
To find the empirical formula mass in g/mol
Formula mass:
Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.
**Note:
if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol
So,
As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O
Then Its empirical formula mass will be
CH₃O
Atomic Mass of C = 12
Atomic Mass of H = 3
Atomic Mass of O = 16
Total Molar mass of CH₃O
CH₃O = 12 + 3(1) + 16
CH₃O = 12 + 3 + 16
CH₃O = 31 g/mol
Isotope 1: 89.905 * 51.45 = 4625.61225 / 100 = 46.2561225
Isotope 2: 90.906 * 11.22 = 1019.96532 / 100 = 10.1996532
Isotope 3: 91.905 * 17.15 = 1576.17175 / 100 = 15.7617075
Isotope 4: 93.906 * 17.38 = 1632.08628 / 100 = 16.3208628
Isotope 5: 95.908 * 2.08 = 268.5424 / 100 = 2.685424
46.2561225 + 10.1996532 + 15.7617075 + 16.3208628 + 2.685424 = 91.22377
actual mass Zr = about 91.22
Answer:
Percentage of oxygen = 30%
Percentage of carbon = 30%
Percentage of hydrogen = 40%
Explanation:
Formula:
Percentage of element = given amount / total amount × 100
Given compound:
C₆H₈O₆
Number of atoms of carbon = 6
Number of atoms of hydrogen = 8
Number of atoms of oxygen = 6
Total number of atoms = 20
Percentage of carbon = 6/20 × 100
Percentage of carbon = 30%
Percentage of Hydrogen = 8/20 × 100
Percentage of Hydrogen = 40%
Percentage of oxygen = 6/20 × 100
Percentage of oxygen = 30%
<h3>
Answer:</h3>
16.7 g H₂O
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
