Long term because if you leave something out to be weathered then it can’t be unweathered because of the drastic change of the object.
I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon
Answer:
Explanation:
50 * 10^100 is 50 googols in scientific notation... but
come on, who would write that in standard form...
Answer:
49.5J/°C
Explanation:
The hot water lost some energy that is gained for cold water and the calorimeter.
The equation is:
Q(Hot water) = Q(Cold water) + Q(Calorimeter)
<em>Where:</em>
Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J
Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J
That means the heat gained by the calorimeter is
Q(Calorimeter) = 4794J - 3834J = 960J
The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:
59.4°C-40°C = 19.4°C
And calorimeter constant is:
960J/19.4°C =
<h3>49.5J/°C</h3>
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