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IrinaK [193]
3 years ago
8

Kinetic and potential energy are two types of energy that relate too

Chemistry
1 answer:
Marysya12 [62]3 years ago
7 0

Answer:

Option B. A tractor–trailer traveling at 80 kph.

Explanation:

Kinetic energy can be defined as the energy possessed by a body in motion. Mathematically, it is expressed as:

K.E = ½mv²

Where:

K.E is the kinetic energy.

m is the mass of the object.

v is the velocity of the object.

From the equation, K.E = ½mv²,

We can say that the kinetic energy (K.E), is directly proportional to both the mass (m) and square of the velocity (v). This implies that the greater the mass of an object, the greater the kinetic energy and the smaller the mass, the smaller the kinetic energy.

Now, considering the options given in question above, it is evident that the tractor–trailer has a greater mass than the car, cheetah and motor cycle. Hence, the tractor–trailer will have a greater kinetic energy even though they are traveling with the same velocity.

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blagie [28]

Answer:

if i had to guess it would be D

Explanation:

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What term describes the process of lowering the boiling point of a hydrocarbon by exposing it to different environmental treatme
IgorLugansk [536]

Answer:

Cracking.

Explanation:

A chemical reaction can be defined as a reaction in which two or more atoms of a chemical element react to form a chemical compound. An example of a chemical reaction involving hydrocarbons is cracking.

Hydrocarbon can be defined as an organic compound that comprises of hydrogen and carbon only.

In Organic chemistry, cracking refers to the process of lowering the boiling point of a heavy, complex or long-chain hydrocarbon such as kerogens by exposing it to different environmental treatments such as hydrogen enriched catalysts, pressure or high temperatures, in order to produce smaller, lighter and more useful molecules (alkanes and alkenes) such as gasoline, diesel fuel, etc.

8 0
3 years ago
Which of the following properties of water is due primarily to the uneven distribution of charge between the hydrogen and oxygen
Blizzard [7]
<span>ability to dissolve ionic substances is the answer</span>
4 0
3 years ago
Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,
djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

Relation of K_p\text{ with }K_c is given by the formula:

K_p=K_c(RT)^{\Delta n_g}

where,

K_p = equilibrium constant in terms of partial pressure = ?

K_c = equilibrium constant in terms of concentration = 6.5\times 10^4

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 35^oC=[35+273]K=308K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43

K_p is the constant of a certain reaction at equilibrium while Q_p is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of Q_p for above equation follows:

Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}

We are given:

p_{NOCl}=1.76atm

p_{NO}=1.01atm

p_{Cl_2}=0.42atm

Putting values in above equation, we get:

Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23

We are given:

K_p=1583.43

There are 3 conditions:

  • When K_{p}>Q_p; the reaction is product favored.
  • When K_{p}; the reaction is reactant favored.
  • When K_{p}=Q_p; the reaction is in equilibrium

As, K_p>Q_p, the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

4 0
3 years ago
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc
Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is C_5H_7N

(b) The empirical formula for the given compound is C_4H_5N_2O

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = \frac{6.17}{1.23}=5.01\approx 5

For Hydrogen  = \frac{8.70}{1.23}=7.07\approx 7

For Nitrogen = \frac{1.23}{1.23}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is C_5H_7N_1=C_5H_7N

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = \frac{4.12}{1.03}=4

For Hydrogen  = \frac{5.19}{1.03}=5.03\approx 5

For Nitrogen = \frac{2.06}{1.03}=2

For Nitrogen = \frac{1.03}{1.03}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is C_4H_5N_2O_1=C_4H_5N_2O

6 0
3 years ago
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