Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
An example for gaining potential energy would be: A glass bottle on the top of a high shelf would have more high potential energy than a glass bottle on the middle or bottom shelf because it has a long way or more farther to fall down or brake.
Explanation:
Remember Potential Energy is the restored energy of an object has.
I hope this helps you!
Answer:
The behavior of molecules in different phases of matter represents a balance between the kinetic energies of the molecules and the attractive forces between them. All molecules are attracted to each other. The molecules are in the solid-state. At higher temperatures, the kinetic energy of the molecules is higher.
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
