Answer:
The answer to this is
The velocity of the 27.3Kg marble after collision is = 16.24 cm/s
Explanation:
To solve the question, let us list out the given variables and their values
Mass of first marble m1 = 27.3g
Velocity of the first marble v1 = 21.0 cm/s
Mass of second marble m2 = 11.7g
Velocity of the second marble v2 = 12.6 cm/s
After collision va1 = unknown and va2 = 23.7 cm/s
From Newton's second law of motion, force = rate of change of momentum produced
Hence m1v1 + m2v2 = m1va1 + m2va2 or
va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s
The velocity of the 27.3Kg marble after collision is = 16.24 cm/s
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.
Answer:
Option D is correct = 8.12 grams of NaCl
Explanation:
Given data:
Moles of sodium chloride = 0.14 mol
Mass of sodium chloride = ?
Solution:
Formula:
Number of moles = mass of NaCl / Molar mass of NaCl
Molar mass of NaCl = 58 g/mol
Now we will put the values in formula.
0.14 mol = Mass of NaCl / 58 g/mol
Mass of NaCl = 0.14 mol × 58 g/mol
Mass of NaCl = 8.12 g of NaCl
Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.
