Answer:
CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH > ClCH₂COOH
Explanation:
Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.
Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.
- The closer the substituent is to the carboxyl group, the greater is its effect.
- The more substituents, the greater the effect.
- The effect tails off rapidly and is almost zero after about three C-C bonds.
CH₃CH₂-CH₂COOH — EDG — weakest — pKₐ = 4.82
CH₃-CH₂COOH — reference — pKₐ = 4.75
ClCH₂-CH₂COOH — EWG on β-carbon— stronger — pKₐ = 4.00
ClCH₂COOH — EWG on α-carbon — strongest — pKₐ = 2.87
<h2>Question:- </h2>
A solution has a pH of 5.4, the determination of [H+].
<h2>Given :- </h2>
- pH:- 5.4
- pH = - log[H+]
<h2>To find :- concentration of H+</h2>
<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>
<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>
Take negative to other side
-pH = log H+
multiple Antilog on both side
(Antilog and log cancel each other )
Antilog (-pH) = [ H+ ]
New Formula :- Antilog (-pH) = [+H]
Now put the values of pH in new formula
Antilog (-5.4) = [+H]
we can write -5.4 as (-6+0.6) just to solve Antilog
Antilog ( -6+0.6 ) = [+H]
Antilog (-6) × Antilog (0.6) = [+H]
put the value in equation
![{10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]](https://tex.z-dn.net/?f=%20%7B10%7D%5E%7B%20-%206%7D%20%20%20%5Ctimes%204%20%3D%20%5BH%2B%5D%20%5C%5C%204%20%5Ctimes%20%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%3D%20%5BH%2B%5D)
Answer:
= 46.06 grams
Explanation:
P=cRT
25 = c (0.0821)(298)
concentration = c = 1.021 M
number of moles = (1.021)(0.75) = 0.7664 moles
mass = (0.7664)(60.10)=40.06 grams
Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
A neutral atom of potassium has 19 electrons.
