<h3>
Answer:</h3>
3.0 × 10²³ molecules AgNO₃
<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>
- Reading a Periodic Table
- Writing Compounds
- 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>
85 g AgNO₃ (silver nitrate)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 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>
3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃
Answer:
The answer to your question is: 25 g of PbCl2
Explanation:
Data
NaCl = 25 g
PbCl₂ = ?
Pb(NO3)2 + 2 NaCl ⇒ PbCl2 + NaNO3
MW NaCl = 58.5 g
MW PbCl2 = 277 g
2(58.5 g) of NaCl ------------------------ 277 g of PbCl2
25 g of NaCl ------------------------ x
x = (25 x 277) / 117
x = 25 g
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
Answer: Kb = 3.15 × 10 ⁻⁴
Explanation:
This is how you calculate Kb for this reaction.
1) Equilibrium equation:
CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻
2) Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂] ↔ all the spieces in equilibrium
3) From the stoichiometry [CH₃NH₃⁺] = [OH⁻]
Then, Kb = [OH⁻] [OH⁻] / [CH₃NH₂] = [OH⁻]² / [CH₃NH₂]
4) You get [OH⁻] from the pH in this way:
pOH + pOH = 14 ⇒ pOH = 14 - pH = 14 - 11.40 = 2.60
pOH = - log [OH⁻] = 2.60 ⇒ [OH⁻] = 10^(-2.6) = 0.002512
5) [CH₃NH₂] in equilibrium is given: 0.0200M
6) Now compute:
Kb = (0.002512)² / 0.0200 = 3.15 × 10 ⁻⁴
