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.
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Answer:
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Answer:
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Explanation:
To determine limiting reactant divide mole quantities of reactants by the respective coefficient in the balanced equation. The smaller value is the limiting reactant.
P₄ + 5O₂ => 2P₂O₅
12/1 = 12 15/5 = 3
O₂ is the limiting reactant. P₄ will be in excess when rxn stops.
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
