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3 years ago

Topic experiment of rate of reaction. answer with explanations. don't answer bad things or wrong you would be reported.

Chemistry

1 answer:

Zigmanuir [339]3 years ago

3 0

Answer:

(I dont know how to write small numbers)

There is no gas bubbles if you have (+C6H8O7)

Explanation:

1: answer is 1 and 3

2: answer is 2 and 4

3: answer is 3

4: answer is 1

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3 years ago

The type of chemical bond that involves the transfer of electrons between atoms is a(n) bond.

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An ionic bond occurs when an atom transfers or takes Electrons from another atom.

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3 years ago

Read 2 more answers

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

At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution

IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = $x_{A}\times P_{A}^{0}$

vapour pressure of a component (B) in solution = $x_{B}\times P_{B}^{0}$

Where $x_{A},x_{B}$ are mole fraction of component A and B in solution respectively

$P_{A}^{0},P_{B}^{0}$ are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = $0.340\times 745 torr$

= 253 torr

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3 years ago

What’s the difference between grazing and true predation

Mila [183]

Answer:

For terrestrial animals, grazing is normally distinguished from browsing in that grazing is eating grass or forbs, and browsing is eating woody twigs and leaves from trees and shrubs. Grazing differs from true predation because the organism being grazed upon is not generally killed.

Explanation:

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3 years ago

Topic experiment of rate of reaction. answer with explanations. don't answer bad things or wrong you would be reported.​

Topic experiment of rate of reaction. answer with explanations. don't answer bad things or wrong you would be reported.