Answer:
Amino >Methoxy > Acetamido
Explanation:
Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.
The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.
The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.
Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.
Therefore, decreasing order will be as follows:
Amino >Methoxy > Acetamido
Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
Answer:-
0.91% is the students % of error
Explanation: -
Accepted value= 12.11 grams
Measured value = 12.22 grams
Error = 12.22-12.11 = 0.11 grams
Percentage error = 
x100
= 0.91 %
Thus 0.91% is the students % of error
Copper nitrate and nitric oxide are produced in this reaction.
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
