A ball of mass 0.2 kg is dropped from a height of 10 m. How much mechanical energy does it have right before it hits the ground?
(Assume there is no air resistance.) Acceleration due to gravity is g 9.8 m/s2.
(PLEASE HELP NEED ANSWERS ASAP)
(PLEASE HELP NEED ANSWERS ASAP)
1 answer:
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The solubility equilibrium of :
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.
Answer : The change in entropy is
Explanation :
Formula used :
where,
= change in entropy = ?
m = mass of water = 1.00 kg
= heat of vaporization of water =
T = temperature =
Now put all the given values in the above formula, we get:
Therefore, the change in entropy is