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: CF4
Explanation:
Calculate the molar mass of each compound. Divide the molar mass of Carbon by the molar mass of each compound, then multiply the answer by 100 to get the percentage.
CF4= 12+(19X 4)
=12+76= 88 g/mol
%C= 12/88 x 100= 13.64%
CO2= 12+(16 X 2)
12+32= 44 G/MOL
%C= 12/44 x 100= 27.3%
CH4= 12+ (1 X4)
=12+4
=16 G/MOL
%C= 12/16 X 100= 75%
C204
(12X2) + (16X4)
24+64
= 88 g/mol
%C= 24/88 x 100
= 27.3%
Answer:
D. 12
Explanation:
The pH is a measure of the acidity of a solution. The pH indicates the concentration of hydronium ions [H3O +] present in a solution;
pH = - log [H3O+]
So
pH= - log [7.8 × 10−13 M]
finally
pH= 12
Answer:
38 kg/m³
0.038 g/mL
Explanation:
Volume of a cube is the side length cubed.
V = s³
Given s = 0.65 m:
V = (0.65 m)³
V ≈ 0.275 m³
The mass is 10.5 kg. The density is the mass divided by volume:
ρ = (10.5 kg) / (0.275 m³)
ρ ≈ 38 kg/m³
Or:
ρ ≈ 0.038 g/mL
Answer:
The answer to your question is: 516 g of water
Explanation:
2 C₇H₁₄ + 21 O₂ → 14 CO₂ + 14 H₂O
8 moles 43 moles
Process
1.- Find the limiting reactant
Ratio theoretical O₂ / C₇H₁₄ = 21 / 2 = 10.5
Ratio experimental O₂ / C₇H₁₄ = 43 / 8 = 5.3
As the ratio diminishes the limiting reactant is O₂.
2.- Calculate the moles of water
21 moles of O₂ ------------- 14 moles of water
43 moles of O₂ ------------ x
x = (43 x 14) / 21
x = 28.67 moles of water
3.- Calculate the grams of water
Molecular mass of water= 18g
18 g ------------------------ 1 mol of water
x ------------------------ 28.67 moles of water
x = (28.67 x 18) / 1
x = 516 g of water
