Answer:
d
Explanation:
Carbohydrates are compounds containing carbon, hydrogen, and oxygen. Therefore, a is true.
An empirical formula is the simplest ratio of atoms present in a compound. Therefore, C2H4O2 and C3H6O3, (if you simplified them like you would a fraction) would be CH2O. Therefore b is correct,
They also have the same % composition, with a ratio of 1 carbon : 2 hydrogen : 1 oxygen. Therefore, c is correct.
Since a, b and c are all correct, the answer is d, all of the above are true.
Answer:
See explanation
Explanation:
The compound ClO2 has 19 valence electrons. ClO2 is a bent molecule with tetrahedral electron pair geometry but has two lone pairs of electrons. This is indicated by the presence of four electron pairs on the outermost shell of the central atom.
The molecule has an odd number of valence electrons, hence, it is generally regarded as a paramagnetic radical. None of the proposed Lewis structures for the molecule is satisfactory because none of them obeys the octet rule.
From the images attached, one can easily see that the electron dots around the oxygen and chlorine atoms does not satisfy the octet rule in all the resonance structures shown.
1/2 - 3/7 = 7/14 - 6/14 = 1/14
The pH of the monoprotic weak acid is 2.79.
<h3>What are weak acids?</h3>
The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.
The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c (monoprotic acid) = 0.33 M.
Ka = 1.2·10⁻⁶
[A⁻] = [H⁺] = x
[HA] = 0.33 M - x
Ka = [A⁻]·[H⁺] / [HA]
2. 6 × 10⁻⁶ = x² / (0.33 M - x)
Solve quadratic equation: [H⁺] = 0.000524 M.
pH = -log[H⁺]
pH = -log(0.000524 M)
pH = 2.79
Thus, the pH of the monoprotic weak acid is 2.79
To learn more about weak acids, refer to the below link:
brainly.com/question/13032224
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Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
