Lightning heats the surrounding air to A. 300°C. B. 30,000°C. C. 3,000°C. D. 300,000°C.
2 answers:
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<h3>Answer:</h3>
3.01 × 10²⁵ molecules 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
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Step 1: Define</u>
50.0 mol H₂O
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
<u /> = 3.011 × 10²⁵ molecules H₂O
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O
Answer:
The pH of saturated solution of the quinine is 10.05
Explanation:
Quinine (Q) is C20H24N2O2 has a molar mass of 324.4 g/mol
Q can behave as a weak base. Kb and pKb can be calculated for weak bases
pKb1 is to be considered when solving the question.
pKb1 = 5.1
Step 1 : Calculate the Kb of Quinine
pKb1 = - log [kb]
5.1 = - log [kb]
take Antilog of both side
[kb] = 7.94 x 10∧-6
Step 2: Calculate the concentration of saturated solution of Q in mol/dm3
From the question, 1900 ml of solution contains 1 g of Q
Therefore, 1000 ml of solution will contain........... x g of Q
x = 1000 /1900
x = 0.526 g in 1 dm3
In calculating concentration in mol/dm3,
Concentration in mol/dm3 = concentration in g /dm3 divided by molar mass
Molar mass of Q = 324.4
Concentration in mol/dm = 0.526 /324.4
= 0.0016 mol/dm3
Step 3: Calculating the Concentration of OH-
At Equilibrium, Kb = x² / 0.0016
7.94 x 10∧-6 = x² / 0.0016
x = √ 0.0016 × 7.94 x 10∧-6
x = 1. 127 × 10∧-4 mol/dm3
The concentration of OH- = 1. 127 × 10∧-4 mol/dm
Step 4: Calculating the pH of Quinine
Recall, pOH = - log [OH-]
pOH = - log [1. 127 × 10∧-4]
pOH = 3.948
Also recall that pH + pOH = 14
pH = 14 - 3.948
pH = 10.05
