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adell [148]
3 years ago
6

What would a chemist most likely study about a car? 1- The amount of rotations of the wheel that occur in 1 mile. 2- The reactio

n of gasoline combustion in the engine. 3- The accuracy of the speedometer. 4- The strength of the car frame.
Chemistry
1 answer:
NikAS [45]3 years ago
4 0

Answer: B: the reaction of gasoline combustion in the engine

Explanation:

I just took the quiz

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This is science
irakobra [83]

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6 0
3 years ago
Which of the following objects would take you the greatest amount of force to accelerate?
horsena [70]

Complete question is;

Which of the following object would take you the greatest amount of force to accelerate.

A) a soccer ball with a mass of 0.5 kg

B) a refrigerator with a mass of 200 kg, C) a bike with a mass of 25 kg

D) a car with a mass of 5,000 kg,

Answer:

D) a car with a mass of 5,000 kg

Explanation:

Formula for force is;

F = ma

Where;

F is force

m is mass

a is acceleration

Now, Force is directly proportional to the acceleration and mass.

Thus, the higher the mass, the greater the force.

Thus, the object that will require the most force is the one that has the highest mass.

Looking at the options, the one with the highest mass is option D.

3 0
3 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
In 1953, who devopled the model that is shown below
PilotLPTM [1.2K]
What model? can you screenshot it or send a link?
7 0
3 years ago
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