Chemistry, 06.04.2021 16:40, rfcordray307

How many protons do you think are dissociable from propionic acid and butyric acid

Answers: 3

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answered: Guest

I’m pretty sure it’s false i’m not sure

answered: Guest

hey there!

mass = 6.75 g

volume = 9 cm³

therefore:

d = mass / volume

d = 6.75 / 9

d = 0.7 g/cm³

answer b

answered: Guest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

answered: Guest

answer : the correct answer is 2) hno₃ and naoh .

arrhenius concept of acid and base :

acid :

according to arrhenius acid are those which release h⁺ or hydronium ion (h₃o⁺) in aqueous solution .

example : hcl +h₂o ↔ h₃o⁺ + cl⁻

hcl is giving h₃o⁺ ion in water , hence it is an arrhenius acid .

base :

according to arrhenius , bases are those which releases oh⁻ ions in aqueous solution .

example : koh (aq) ↔ k⁺ + oh⁻

since koh is producing oh⁻ ions hence , it is an arrhenius base .

checking the options :

1) ch₃oh and naoh :

ch₃oh is known as methanol , an organic compound , which is neutral compound . the dissociation of ch₃oh is negligible . it is very very weak acid (ka = 3.2 * 10 ⁻¹² ) .

naoh being ionic gives na⁺ and oh⁻ ions in aqueous solution as :

naoh ↔ na⁺ + oh⁻ . hence it is an arrhenius base

it can be concluded that ch₃oh and naoh is not arrhenius acid and base.

2) hno₃ and naoh :

hno₃ is an ionic compound which gives ions h⁺ and no₃⁻ ion in aqueous solution as:

hno₃ + h₂o ↔ h₃o⁺ + no₃⁻

since it is giving h⁺ ion . hno₃ is arrhenius acid .

naoh being ionic gives na⁺ and oh⁻ ions in aqueous solution as :

naoh ↔ na⁺ + oh⁻

. hence it is an arrhenius base .

it can be said that hno₃ and naoh are arrhenius acid and base.

3) ch₃oh and hcl