Answer:
i think that the answer is that it would decrease
Explanation:
hope this helps
sorry if the answer is wrong
Answer:
It will be reported too low.
Explanation:
To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.
Qmetal + Qwater = 0
Qmetal = -Qwater
The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:
m_metal*s*ΔT_metal = - m_water *c*ΔT_water
s = -m_water *c*ΔT_water / m_metal*ΔT_metal
So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.
True because it is warmer closer to the equator
The correct option is D.
During peptide formation, when two amino acids come together, a hydrogen atom and two molecules of oxygen are released in form of water molecules. The amino acid that present the carboxyl group to the reaction loses an hydroxyl group while the amino acid that present the amino group to the reaction loses a hydrogen.
