A nuclide of phosphorus 3015P decays into a nuclide of silicon 3014Si. In order to satisfy charge conservation and lepton number
1 answer:
In order to satisfy charge conservation and lepton number conservation the other products must be neutron.
<h3>What is conservation of mass?</h3>
The principle of conservation of mass states that, the sum of the initial mass of reactants must be equal to final mass of the products.
The balanced reaction of radioactive decay of phosphorous shows conservation of mass.
Thus, in order to satisfy charge conservation and lepton number conservation the other products must be neutron.
Learn more about radioactive decay here: brainly.com/question/1383030
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Answer:- 2.92 kJ of heat is released.
Solution:- We have water at 100 degree C and it's going to be cool to 15 degree C.
So, change in temperature, = 15 - 100 = -85 degree C
mass of water, m = 8.2 g
specific heat of water, c =
The equation used for solving this type of problems is:
Let's plug in the values in the equation and solve it for q which is the heat energy:
q = (8.2)(4.184)(-85)
q = -2916.248 J
They want answer in kJ. So, let's convert J to kJ and for this we divide by 1000.
q = -2.92 kJ
Negative sign indicates the heat is released. So, in the above process of coiling of water, 2.92 kJ of heat is released.