Which of the following pairs of elements are chemically most similar to each other?

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

Which of the following is likely to have the lowest viscosity? hot oil room temperature oil room temperature water below room te

zzz [600]

Viscosity, is a state of matter where it resists flow or may be in thick semifluid form because of friction acting inside the matter.In this case, hot oil room temperature would lessen the viscosity of the oil (animal oil). Thank you for your question. Please don't hesitate to ask in Brainly your queries.

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4 years ago

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Tonicity (how much salt is in it)

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3 years ago

You are given solutions of hcl and naoh and must determine their concentrations. you use 27.5 ml of naoh to titrate 100.0 ml of

Dafna1 [17]

1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.

2) Equation:

2Na OH + H2SO4 --> Na2 SO4 + 2H2O

3) molar ratios

2 mol NaOH : 1 mol H2SO4

4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution

M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4

5) Number of moles of NaOH

2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH

6) Concentration of the solution of NaOH

M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M

7) Standardize the solution of HCl

Chemical reaction:

NaOH + HCl --> NaCl + H2O

8) Molar ratios

1 mol NaOH : 1 mol HCl

9) Number of moles of NaOH in 27.5 ml

M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH

10) Number of moles of HCl

1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl

11) Concentration of the solution of HCl

M = n / V = 0.01169 mol / 0.100 l = 0.1169 M

Rounded to 3 significant figures = 0.117 M

Answers:

[NaOH] = 0.425 M
[HCl] = 0.117 M

3 0

3 years ago