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kifflom [539]
1 year ago
11

If you feel warm after exercising, have you increased the internal energy of your body? Explain.

Chemistry
1 answer:
elena-s [515]1 year ago
6 0

If we feel warm after exercising, it means that the temperature of the surroundings has increased. Therefore, heat is released from our body (energy transferred from the system to the surroundings) which means the internal energy of our body is decreased after exercising.

internal energy U is the sum of the kinetic energy brought about by the motion of molecules and the potential energy brought about by the vibrational motion and electric energy of atoms inside molecules in a system or a body with clearly defined limits. The energy contained in every chemical link is often referred to as internal energy. From a microscopic perspective, the internal energy can take on a variety of shapes. For any substance or chemical attraction between molecules.

Internal energy is a significant amount and a state function of a system. Specific internal energy, which is internal energy per mass of the substance in question, is a very intense thermodynamic characteristic that is often represented by the lowercase letter U. As a result, the J/g would be the SI unit for internal specific energy. The term "molar internal energy" and the unit "J/mol" might be used to describe internal energy that is expressed as a function of the quantity of a substance.

Learn more about internal energy brainly.com/question/11278589

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8 0
2 years ago
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A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec
GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol

  • <u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol

For the given chemical reaction:

Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, 3.14\times 10^{-5} moles of NaI will react with = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 3.14\times 10^{-5} moles of NaI will produce = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

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Answer:

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Explanation:

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