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8_murik_8 [283]
2 years ago
15

The law of reflection states that the angle of incidence and the angle of reflection are always

Chemistry
2 answers:
igomit [66]2 years ago
8 0

Answer:

equal in measure

Explanation:

got it correct on edg

cestrela7 [59]2 years ago
6 0

Answer:

The law of reflection states that the angle of incidence and the angle of reflection are always equal.

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It usually takes Vaughn 12 minutes to drive to school. She drives an average of 50.0 miles per hour. What is the distance from s
lilavasa [31]
<h3>Answer:</h3>

  = 633,600 Inches

<h3>Explanation:</h3>

Speed refers to the rate of change in distance traveled by a body in motion.

It is the ratio of distance to time taken.

It is calculated by;

Speed = Distance ÷time taken

Therefore;

Distance = Speed × time

In this case;

Time taken = 12 minutes ( but 1 hr = 60 mins)

Thus, t = 0.2 hours

Speed = 50.0 miles/hour

Thus;

Distance = 50.0 miles/hr × 0.2 hours

               = 10 miles

But, we are require to give the distance in inches;

1 mile = 63360 Inches

Therefore;

10 miles = 10 × 63360 Inches

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6 0
3 years ago
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

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3 years ago
Fill in the blanks to determine the number of protons and electrons in an oxygen ion. (Consult the periodic table to determine t
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The answer would be 2
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C!!!! The third one!
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What is the difference between endothermic reactions and exothermic reactions? (Please describe both types of chemical reactions
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Answer:

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