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Basile [38]
3 years ago
8

When a collision occurs between two reactant particles that, between them, have the required minimum kinetic energy, or activati

on energy, a product does not always form. Which of the following reasons explains this?
a.low temperature
b.small surface area
c.unfavorable geometry
d.low concentration
Chemistry
2 answers:
erastovalidia [21]3 years ago
7 0

c.unfavorable geometry


iris [78.8K]3 years ago
6 0

Answer:  C.  Unfavorable geometry

Explanation: When collision occurs between two reactants in order to make a reaction possible there are 3 factors which are responsible.

a) Orientation factor

b) Energy factor

c) rate of collision

Thus out of the given options, unfavorable geometry is the correct one as temperature and concentration as well as surface area will have very little effect on the reaction.

If the geometry of the reactant is not complementary then the reaction would not lead successfully.

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Statement 3 is false because the kinetic energy of a particle determines its heat and it is impossible for a particle to be completely motionless.

Statement 4 is false because neutrons do not have a charge and make up a large portion of all matter.  

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Process in which water vapor turns to liquid
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Answer: 1045 J of energy was released on cooling the down the water from 20 °C to 10 °C.

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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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