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mr_godi [17]
3 years ago
5

Which statements describe or apply the principal of conservation of energy? Select all that apply.

Chemistry
2 answers:
laila [671]3 years ago
7 0

Answer:

A, C, and D

Explanation:

The correct options that apply to the principal conservation of energy are A, C, and D.

A is correct because energy can neither be created nor destroyed. However, energy can be transfered from one location to another or be converted from one form to another. <em>Whether transferred to converted, the magnitude remains the same.</em>

C is correct because energy cannot be destroyed but can be transferred or converted. <em>Hence, if a body or a location loses temprature, then the loss is being gained by another body or location.</em>

D is also correct. A closed system is a system that does not exchange matter with its surroundings. <em>Hence, the total energy remains the same within the system. </em>

Harrizon [31]3 years ago
6 0

Answer:

The total energy in a closed system such as Earth stays the same.

Explanation:

The simplest statement of the law of conservation of energy is that energy can neither be created nor destroyed but can be converted from one form to another.

A typical illustration of this principle of conservation of energy is a closed system of which our universe is an important example.

The total amount of energy in the universe is always a constant. Energy lost in one part must be equal to the energy gained on the other part.

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Answer:

option c) 3 is the correct option

Explanation

as we  know that 3rd principal energy level contains 3 sub levels,which are named as s,p and d. These sub levels further contain different numbers of orbitals,

and these sub levels can be termed as  regions of probability of finding an electron, and each orbital may have a maximum number of two electrons in it.

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4 years ago
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Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
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Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

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4 years ago
Choose the ingredients needed for nuclear fusion. Check all that apply. energy helium gas high temperatures hydrogen gas low pre
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high temperatures

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3 years ago
What is the mass of 25 Liters of nitrogen dioxide gas?
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Answer:

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Explanation:

mruno bars is the wae

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3 years ago
A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou
Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

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4 years ago
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