In 0.190 mole of C6H14O, there is 0.190*6 (number of C in one molecule) = 1.140 mole of C atoms. The total number of C atoms = 1.14 * 6*
(atoms of C in one mole) = 6.84*
atoms.
Answer:
The mass percent of potassium is 39%
Option C is correct
Explanation:
Step 1: Data given
Atomic mass of K = 39.10 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate molar mass of KHCO3
Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0
Molar mass KHCO3 = 100.12 g/mol
Step 3: Calculate mass percent of potassium (K)
%K = (atomic mass of K / molar mass of KHCO3) * 100%
%K = (39.10 / 100.12) * 100%
%K = 39.05 %
The mass percent of potassium is 39%
Option C is correct
First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹
The answer is C
Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.
NH₂-CH₂-COOH + HNH-CH₂-COOH → NH₂-CH₂-CO-NH-CH₂-COOH + H₂O
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