To find the mass of reagent to be added in the reaction, we need to determine the pH of the buffer. Using the following equation
pH= pKa + log ([NH3] / [NH4+])
= 9.26 + LOG ( 0.3 / 0.3)
= 9.26
[h+]= 5.5x10^-10 M, you need to increase that concentration for a ph of 8.6, [H+] needs to be 2.51x10^-9M
2.51x10^-9 moles - 5.5x10^10 moles = 1.96x10^-9 moles
1.96x10^-9 x 36.45g/mole = 7.14x10^-8 g
So the mass of HCI that you should add to the reaction is 7.14x10^-8 g
Answer:
Butan-2-one
Explanation:
1. 1700 cm⁻¹
A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.
2. Triplet-quartet
A triplet-quartet pattern indicates an ethyl group.
The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.
The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.
3. 3H(s) at δ 2.13
This indicates a CH₃ group with no adjacent hydrogen atoms.
It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.
4. Identification
The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.
The compound is butan-2-one.
Evaporation, Condensation, precipitation and collection would be stages
Answer:
in first picture pressure is low and in second picture the pressure is high
2ans in first picture air is
less amount
in second picture air is high amount
The oxidation state of N is
B.+5