During endothermic phase change, the potential energy of the system always increases while the kinetic energy of the system remains constant. The potential energy of the reaction increases because energy is been added to the system from the external environment.
<u>Explanation</u>:
- Those are three distinct methods for demonstrating a specific energy condition of an object. They don't affect one another.
- "Potential Energy" is a relative term showing a release of possible energy to the environment. If we accept its pattern as the overall energy state of a compound, at that point, an endothermic phase change would infer an increase in "potential" as energy is being added to the compound by the system.
- A phase change will display an increase in the kinetic energy at whatever point the compound is transforming from a high density to a low dense phase. The kinetic energy will decrease at whatever point the compound is transforming from a less dense to high dense phase.
The volume of one mole of any gas at STP is 22.4 L. So, at STP, the volume of 2.00 moles of hydrogen gas would be (22.4 L/mol)(2 mol H2) = 44.8 L.
Eukaryotic cells, the theoretical maximum yield of ATP generated per glucose is 36 to 38, depending on how the 2 NADH generated in the cytoplasm during glycolysis enter the mitochondria and whether the resulting yield is 2 or 3 ATP per NADH
Answer:
![\boxed{\text{66.95 g BaSO$_{4}$}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B66.95%20g%20BaSO%24_%7B4%7D%24%7D%7D)
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 261.34 233.39
Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃
m/g: 75.00
1. Moles of Ba(NO₃)₂
![\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Ba%28NO%24_%7B3%7D%29_%7B2%7D%24%7D%20%3D%20%5Ctext%7B75.00%20g%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%7D%7D%7B%5Ctext%7B261.34%20g%7D%7D%20%3D%20%5Ctext%7B0.286%2098%20mol%7D)
2. Moles of BaSO₄
The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂
![\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20BaSO%24_%7B4%7D%24%7D%3D%20%5Ctext%7B0.286%2098%20mol%20Ba%28NO%24_%7B3%7D%29_%7B2%7D%24%20%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20BaSO%24_%7B4%7D%24%7D%7D%7B%5Ctext%7B1%20mol%20Ba%28NO%24_%7B3%7D%29_%7B2%7D%24%7D%7D%20%3D%20%5Ctext%7B0.286%2098%20mol%20BaSO%24_%7B4%7D%24%7D)
3. Mass of BaSO₄
![\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20BaSO%24_%7B4%7D%24%7D%20%3D%20%5Ctext%7B0.286%2098%20mol%20BaSO%24_%7B4%7D%24%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B233.39%20g%20BaSO%24_%7B4%7D%24%7D%7D%7B%5Ctext%7B1%20mol%20BaSO%24_%7B4%7D%24%7D%7D%20%3D%20%5Ctextbf%7B66.98%20g%20BaSO%24_%7B4%7D%24%7D%5C%5C%5C%5C%5Ctext%7BThe%20theoretical%20yield%20of%20barium%20sulfate%20is%20%7D%20%5Cboxed%7B%5Ctextbf%7B66.98%20g%20BaSO%24_%7B4%7D%24%7D%7D)
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