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gavmur [86]
3 years ago
7

In printed form, the Oxford English Dictionary __________________.

Chemistry
1 answer:
inna [77]3 years ago
3 0
Will i thank <span>consists of many volumes
hope this help!!!!!!!!</span>
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ELEMENTS ARE CLASSIFIED AS METALS, NONMETALS, AND METALLOIDS. USE THE PERIODIC TABLE AND IDENTIFY TWO OF EACH CLASSIFICATION.
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Elements are classified into- metals – Nonmetals- Metalloids – Noble gases. State which of A, B, C, D is a:

1) Metallic element

2) Non-metallic element

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A) Is non-malleable, non-ductile and a poor conductor of electricity

B) Has lustre, is malleable and ductile and a good conductor of electricity

C) Is unreactive and inert and present in traces in air

D) Shows properties of both metals and nonmetals

Explanation:

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The partial pressure of argon gas, making up 40% of a mixture is 325 torr. What is the total pressure of the mixture in kPa?
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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

5 0
3 years ago
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