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3 years ago

Draw all constitutional isomers with molecular formula c3h8o, and rank them in terms of increasing acidity:

Chemistry

1 answer:

Nookie1986 [14]3 years ago

8 0

There are only three constitutional isomers (see the image below).

Ethyl methyl ether is the least acidic because it has only C-H bonds. However, it should be more acidic than an alkane (pKₐ ≈ 50) because of the electron withdrawing effect of the O atom. I would expect it to have pKₐ ≈ 30.

The alcohols are more acidic because the O in the O- H bond is highly electron withdrawing. Alcohols typically have pKₐ ≈ 17.

Alkyl groups are electron-donating. This increases electron density in the OH bond and makes the acid weaker. Propan-2-ol has two methyl groups on the α-carbon, while propan-1-ol has one ethyl group. Thus propan-2-ol should be a slightly weaker acid than propan-1-ol.

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Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other h

11Alexandr11 [23.1K]

Complete question:

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.

benzene boiling point=80.1 Kb=2.53

carbon tetrachloride boiling point=76.8 Kb=5.03

Answer:

m = 1.32 mol/kg

Boiling point: 83.4°C

Explanation:

When a nonvolatile solute is added to a pure solvent, the boiling point of the solvent increases, a phenomenon called ebullioscopy. This happens because of the interactions between the solute and the solvent. The temperature variation (new boiling point - normal boiling point) can be calculated by:

ΔT = m*Kb*i

Where m is the molal concentration (moles o solute/mass of solvent in kg), Kb is the ebullioscopy constant of the solvent, and i is the van't Hoff factor, which indicates how much of the solute dissociates. Let's assume that i is equal in both solvents and equal to 1 (the solvent dissociates completely)

Calling the new boiling point as Tb, for benzene:

Tb - 80.1 = m*2.53*1

Tb = 2.53m + 80.1

For carbon tetrachloride:

Tb - 76.8 = m*5.03*1

Tb = 5.03m + 76.8

Because Tb and m are equal for both:

5.03m + 76.8 = 2.53m + 80.1

2.5m = 3.3

m = 1.32 mol/kg

So, substituting m in any of the equations (choosing the first):

Tb = 2.53 * 1.32 + 80.1

Tb = 83.4°C

3 0

3 years ago

Read 2 more answers

The conversion of methyl isonitrile to acetonitrile in the gas phase at 250 °C CH3CN(g) is first order in CH3NC with a rate cons

nadya68 [22]

Answer: The concentration of $CH_3NC$ will be $1.56\times 10^{-2}M$ after 416 seconds have passed.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.00\times 10^{-3}s^{-1}$

t = age of sample = ?

a = let initial amount of the reactant = $5.45\times 10^{-2}M$

a - x = amount left after decay process = $1.56\times 10^{-2}M$

$t=\frac{2.303}{3.00\times 10^{-3}}\log\frac{5.45\times 10^{-2}}{1.56\times 10^{-2}}$

$t=416s$

The concentration of $CH_3NC$ will be $1.56\times 10^{-2}M$ after 416 seconds have passed.

5 0

3 years ago

A solution of pH 5 is diluted 100 times. Find the pH of the resulting solution.

liq [111]

First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:

pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M

M1V1 = M2V2
1 x 10^-5 M (V1) = M2(100V1)
M2 = 1 x 10^-7

pH = -log[1 x 10^-7]
pH = 7

3 0

2 years ago

Read 2 more answers

Waste water from nuclear power plants is generally

FrozenT [24]

Answer:

waste water from nuclear power plant is

generally very harmful for our environment

and for everyone

hope it will help

4 0

2 years ago