What is chemical energy? (2 points)a
Energy that is stored or released during a chemical reaction

b
Energy that is destroyed during a physical change

c
Energy that is released during a phase change

d
Energy that is created when a substance is being burned

Answers

Answer 1
Answer:

Final answer:

Chemical energy is the energy that is stored or released during a chemical reaction. It exists within the chemical bonds of molecules and can be converted into other forms of energy.

Explanation:

Chemical energy refers to the energy that is stored or released during a chemical reaction. It is a form of potential energy that exists within the chemical bonds of molecules. When a chemical reaction takes place, these bonds are broken or formed, resulting in the release or absorption of energy.

For example, when a fuel such as gasoline is burned, the chemical energy stored in the hydrocarbon molecules is converted into heat and light energy. Similarly, during photosynthesis, plants convert sunlight into chemical energy in the form of glucose.

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Answer 2
Answer:

Explanation:

Chemical energy

Chemical energy is energy stored in the bonds of chemical compounds, like atoms and molecules. This energy is released when a chemical reaction takes place.

Usually, once chemical energy has been released from a substance, that substance is transformed into a completely new substance.


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1.81 g H2 is allowed to react with 10.2 g N2, producing 2.19 g NH3.What is the theoretical yield in grams for this reaction under the given conditions?3H2(g)+N2(g)→2NH3(g)

Answers

The theoretical yield : = 10.251 g

Further explanation

Given

Reaction

3H₂(g)+N₂(g)→2NH₃(g)

1.81 g H₂

10.2 g N₂

2.19 g NH₃

Required

The theoretical yield

Solution

Find limiting reactant :

H₂ : 1.81 g : 2 g/mol = 0.905 mol

N₂ : 10.2 g : 28 g/mol = 0.364 mol

mol : coefficient

H₂ = 0.905 : 3 = 0.302

N₂ = 0.364 : 1 = 0.364

H₂ as a limiting reactant(smaller ratio)

Moles NH₃ based on H₂, so mol NH₃ :

= 2/3 x mol H₂

= 2/3 x 0.905

=0.603

Mass NH₃ :

= mol x MW

=0.603 x 17 g/mol

= 10.251 g

What is colloidal solutions

Answers

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

Answer:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid.

The osmotic pressure, π, of a solution of glucose is 132 atm . find the molarity of the solution at 298 k.

Answers

The formula for osmotic pressure is:

\Pi = iMRT

where \Pi is osmotic pressure, i is van't Hoff's factor, M molarity, R is Ideal gas constant, and T is Temperature.

\Pi = 132 atm

The van't Hoff's factor for glucose, i = 1

R = 0.08206 Latmmol^(-1)K^(-1)

T = 298 K

Substituting the values in the above equation we get,

132 atm = 1* M* 0.08206 Latmmol^(-1)K^(-1)* 298

M = (132 atm)/(1* 0.08206 Latmmol^(-1)K^(-1)* 298) = 5.4797 molL^(-1) \simeq 5.48 molL^(-1)

So, the molarity of the solution is 5.48 molL^(-1).

What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl

Answers

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial  molarity =  1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ =  0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

Final answer:

To prepare 197.4 mL of 0.456 M HCl from 1.27 M HCl, you need 71.03 mL of 1.27 M HCl.

Explanation:

The subject of this problem involves using the concept of molarity in Chemistry. We can use a simple formula for dilution, M1V1 = M2V2, to find the volume. Here M1 (1.27 M) is the molarity of stock HCl, V1 is the required volume, M2 (0.456 M) is the desired molarity, and V2 (197.4 mL) is the volume of the solution. Solving for V1, we get V1 = M2V2 / M1 = (0.456 M * 197.4 mL) / 1.27 M = 71.03 mL. Therefore, 71.03 mL of 1.27 M HCl is needed to prepare 197.4 mL of 0.456 M HCl.

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I need help!! ASAP please..

Answers

Answer:

164 g

Explanation:

Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()

When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.

1) Calculate q for the reaction. You must show your work.

2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.

3) Calculate ΔH for the reaction in kJ/mol. You must show your work.

Answers

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released  by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50  x 4.18 x ( 22 - 16.5 )  ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )  

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH =  18.392 kJ / mol