CuCl2 Determine the name of this compound according to IUPAC nomenclauture. A) copper chloride B) copper (I) chloride C) copper
2 answers:
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Answer:
b.) 25 ml of aluminum hydroxide
Explanation:
For this question, we have to assume that we have the <u>same concentration</u> for all the solutions, for example, <u>1 M</u>. Additionally, we have to take into account the <u>ionization reaction</u> for each species:
a) <u>we have to ions</u>
b) <u>we have fourth ions</u>
c) <u>we have two ions</u>
d) <u>we have one ion</u>
If we have the same volume and the same concentration the variables that will help us to answer the question would be the n<u>umber of ions.</u> If we have <u>more ions we will have more particles dissolved</u>. Therefore the answer would be b) (<u>due to the fourth ions</u>).
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Answer:
a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q
b) the Q-value of this reaction is 5.789 MeV
c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev
Explanation:
a)
The decay equation for the alpha decay is expressed as;
²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q
b)
Calculate the Q-value (in MeV) of this reaction.
Q = Mparent - Mdaughter -Mg
Q = MRa - MRn -Mg
= 224.020202 - 220.011384 - 4.00260305
= 0.00621495 amu
= 5.789 MeV
therefore the Q-value of this reaction is 5.789 MeV
c)
Energy of alpha particle is expressed as;
E∝ = MQ / ( m + M)
now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;
The energy of the alpha particle gives;
E∝ = 220(5.789) / ( 4 + 220) = 5.69 MeV
as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.
Therefore the energy of this alpha is
E∝ = 5.69 - 0.241 = 5.449 Mev
Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev
d)
Sketch of the nuclear decay scheme have been uploaded along side this answer.
