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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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<h3>Answer:</h3>
  • K₂S
  • CaSO₄ (slightly soluble in water)
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<h3>Explanation:</h3>
  • Mg(OH)₂  and CoCO₃ are insoluble in water.
  • Solubility rules are used to determine whether a compound will be soluble or not soluble in aqueous form.
  • K₂S is soluble in aqueous solutions as all salts of Potassium are soluble in water.
  • Calcium (ii) sulfate is slightly soluble in water.
  • Hydroxides are insoluble in water except, KOH, NaOH and NH₄OH.and Barium hydroxide. Therefore, Mg(OH)₂ is an insoluble hydroxide.
  • CoCO₃ is insoluble since carbonates are insoluble except K₂CO₃, Na₂CO₃ and (NH₄)₂CO₃. .
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7 0
3 years ago
A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature rose b
dezoksy [38]

Answer:

872.28 kJ/mol

Explanation:

The heat released is:

ΔH = C*ΔT

where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:

ΔH =  9.47*5.72 = 54.1684kJ

A quantity of 1.922 g of methanol in moles are:

moles = mass / molar mass

moles = 1.992/32.04 = 0.0621 mol

Then the molar heat of combustion of methanol is:

ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol

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