Answer:
Radioisotope therapy is a procedure in which a liquid form of radiation is administered internally through infusion or injection. RIT's ultimate purpose is to treat cancerous cells with minimal damage to the normal surrounding tissue. These therapies are not normally the first approach used to fight a patient's cancer.
Explanation:
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:

We know that,
The relation between the
for an ideal gas are :

As we are given :



Now we have to calculate the entropy change of the gas.


(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Explanation:
As we know
1 liter = 1000 grams
2H2 + O2 --> 2H2O
Weight of H2 molecule = 2.016 g/mol
Weight of water = 18.01 gram /l
2 mole of oxygen react with 2 mole of H2
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Answer:
They all have a certain amount of protons electrons and neutrons.