Sulfuric acid is prepared industrially by means of the response of water with sulfur trioxide which in turn is made by way of a chemical combination of sulfur dioxide and oxygen both by way of the touch system or the chamber system.
H2SO4 (l) H2O (g) + SO3 (g).
The reaction is highly exothermic as an enormous amount of heat is liberated.
The usual approach is to dilute the sulfur trioxide in sulphuric acid. This produces oleum. SO3 (g) + H2SO4 → H2S2O7 (1) Oleum can be in addition diluted in water to acquire concentrated sulphuric acid.
An acid catalyst is added to protonate the carbonyl carbon. How does this catalyze the response, robust acid catalysts catalyze the hydrolysis and transesterification of esters which enables the mechanism with a view to boom the electrophilicity of the carbonyl carbon to assist protonate the carbonyl oxygen.
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Answer:
Sublimation occurs in solids with vapor pressures that exceed atmospheric pressure at or near room temperature.
Explanation:
<h3>
Answer:</h3>
1.69 g Mg₃N₂
<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
- Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg + N₂ → Mg₃N₂
[RxN - Balanced] 3Mg + N₂ → Mg₃N₂
[Given] 1.22 g Mg
[Solve] grams Mg₃N₂
<u>Step 2: Identify Conversions</u>
[RxN] 3 mol Mg → Mg₃N₂
[PT] Molar Mass of Mg - 24.31 g/mol
[PT] Molar Mass of N - 14.01 g/mol
Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol
<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 3 sig figs.</em>
1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂
Answer:
pH before addition of KOH = 4.03
pH after addition of 25 ml KOH = 7.40
pH after addition of 30 ml KOH = 7.57
pH after addition of 40 ml KOH = 8.00
pH after addition of 50 ml KOH = 10.22
pH after addition 0f 60 ml KOH = 12.3
Explanation:
pH of each case in the titration given below
(6) After addition of 60 ml KOH
Since addition of 10 ml extra KOH is added after netralisation point.
Concentration of solution after addition 60 ml KOH is calculated by
M₁V₁ = M₂V₂
or, 0.23 x 10 = (50 + 60)ml x M₂
or M₂ = 0.03 Molar
so, concentration of KOH = 0.03 molar
[OH⁻] = 0.03 molar
pOH = 0.657
pH = 14 - 0.657 = 13.34