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
Answer:
The age of the sample is 4224 years.
Explanation:
Let the age of the sample be t years old.
Initial mass percentage of carbon-14 in an artifact = 100%
Initial mass of carbon-14 in an artifact = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final mass percentage of carbon-14 in an artifact t years = 60%
Final mass of carbon-14 in an artifact = ![[A]=0.06[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D0.06%5BA_o%5D)
Half life of the carbon-14 = 
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%5Ctimes%20t%7D)
![0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}](https://tex.z-dn.net/?f=0.60%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7B5730%20year%7D%5Ctimes%20t%7D)
Solving for t:
t = 4223.71 years ≈ 4224 years
The age of the sample is 4224 years.
Answer : The molecular weight of this compound is 891.10 g/mol
Explanation : Given,
Mass of compound = 12.70 g
Mass of ethanol = 216.5 g
Formula used :
where,
= change in freezing point
= temperature of pure ethanol = 
= temperature of solution = 
= freezing point constant of ethanol = 
i = van't hoff factor = 1 (for non-electrolyte)
m = molality
Now put all the given values in this formula, we get
Therefore, the molecular weight of this compound is 891.10 g/mol
Answer:
a) Se²⁻> S²⁻ > O²
b) Te²⁻ > I- >Cs+
c) Cs+ > Ba²⁺ > Sr²⁺
Explanation:
(a) Se²⁻, S²⁻, O²⁻
In general, ionic radius decreases with increasing positive charge.
As the charge on the ion becomes more positive, there are fewer electrons.
The ion has a smaller radius. In general, ionic radius increases with increasing negative charge.
For ions of the same charge (e.g. in the same group) the size increases as we go down a group in the periodic table
Se²⁻> S²⁻ > O²
(b) Te²⁻, Cs⁺, I⁻
Te²⁻ > I- >Cs+
Te2- hast the biggest size, because of the double negative charge.
Cs+ has the smallest size since it has the most positive charge, compared to Te2- and I-.
(c) Sr²⁺, Ba²⁺, Cs⁺
Cs+ > Ba²⁺ > Sr²⁺
Cs+ has the biggest size, because its more downward (compared to Sr2+) and more to the left (compared) ot Ba2+.
Sr2+ has the smallest size because it's more upwords (compared to Cs+ and Ba2+)
