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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Answer:
The solution to the question is as follows
(a) The rate of ammonia formation = 0.061 M/s
(b) the rate of N₂ consumption = 0.0303 M/s
Explanation:
(a) To solve the question we note that the reaction consists of one mole of N₂ combining with three moles of H₂ to form 2 moles of NH₃
N₂(g) + 3H₂(g) → 2NH₃(g)
The rate of reaction of molecular hydrogen = 0.091 M/s, hence we have
3 moles of H₂ reacts to form 2 moles of NH₃, therefore
0.091 M of H₂ will react to form 2/3 × 0.091 M or 0.061 M of NH₃
Hence the rate of ammonia formation is 0.061 M/s
(b) From the reaction equation we have 3 moles of H₂ and one mole of N₂ being consumed at the same time hence
0.091 M of H₂ is consumed simultaneously with 1/3 × 0.091 M or 0.0303 M of N₂
Therefore the rate of consumption of N₂ = 0.0303 M/s
Answer:
C
Explanation:
the sum of calories is 100. When 50 calories are lost, 50 calories is the remainder.
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
![\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}](https://tex.z-dn.net/?f=%20%5C%5C%20%5Cfrac%7Bdiameter%5C%3Bbiscuit%7D%7Bdiameter%5C%3Batom%28Bi%29%7D)
For this, it is necessary to know the values in meters for any of these diameters:
![\\ 1m = 10^{3}mm = 1e+3mm](https://tex.z-dn.net/?f=%20%5C%5C%201m%20%3D%2010%5E%7B3%7Dmm%20%3D%201e%2B3mm)
![\\ 1m = 10^{12}pm = 1e+12pm](https://tex.z-dn.net/?f=%20%5C%5C%201m%20%3D%2010%5E%7B12%7Dpm%20%3D%201e%2B12pm)
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
![\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m](https://tex.z-dn.net/?f=%20%5C%5C%20320pm%2A%5Cfrac%7B1m%7D%7B10%5E%7B12%7Dpm%7D%20%3D%203.20%2A10%5E%7B-10%7Dm)
<h3>Diameter of a biscuit in meters</h3>
![\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m](https://tex.z-dn.net/?f=%20%5C%5C%2051mm%2A%5Cfrac%7B1%7D%7B10%5E%7B3%7Dmm%7D%20%3D%2051%2A10%5E%7B-3%7Dm%20%3D%205.1%2A10%5E%7B-2%7Dm%20)
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
![\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%20%5Cfrac%7B5.1%2A10%5E%7B-2%7Dm%7D%7B3.20%2A10%5E%7B-10%7Dm%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%20%5Cfrac%7B5.1%7D%7B3.20%7D%5Cfrac%7B10%5E%7B-2%7D%7D%7B10%5E%7B-10%7D%7D%5Cfrac%7Bm%7D%7Bm%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%20%5Cfrac%7B5.1%7D%7B3.20%7D%5Cfrac%7B10%5E%7B-2%7D%7D%7B10%5E%7B-10%7D%7D%5Cfrac%7Bm%7D%7Bm%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%201.5%2A10%5E%7B-2%2B10%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%201.5%2A10%5E%7B8%7D%3D1.5e%2B8)
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
Answer:
Hello There!!
Explanation:
The answer is=>a. Gibbs.
hope this helps,have a great day!!
~Pinky~