<u>Answer:</u>
<u>For a:</u> The isotopic symbol of the above atom will be 
<u>For b:</u> The isotopic symbol of the above atom will be 
<u>For c:</u> the isotopic symbol of the above atom will be 
<u>Explanation:</u>
The isotopic representation of an atom is: 
where,
Z = Atomic number of the atom
A = Mass number of the atom
X = Symbol of the atom
Atomic number of iodine = 53
Mass number of iodine = 131
Symbol of iodine = I
The isotopic symbol of the above atom will be 
- <u>For b:</u> Iridium-192
Atomic number of iridium = 77
Mass number of iridium = 192
Symbol of iridium = Ir
The isotopic symbol of the above atom will be 
- <u>For c:</u> Samarium-153
Atomic number of samarium = 62
Mass number of samarium = 153
Symbol of samarium = Sm
The isotopic symbol of the above atom will be 
Answer:
D.) reducing agent
Explanation:
Oxidized chemicals gain electrons. In order to gain these electrons, another chemical must lose electrons. So, the chemical (which will be oxidized) acts as a reducing agent, causing the other chemical to be reduced and lose electrons.
The answer would be 3.37x10 to exponent 17.
<span>Your final answer would be C4H10O2, which equals 90amu</span>
Answer:
The rate of the reaction increased by a factor of 1012.32
Explanation:
Applying Arrhenius equation
ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)
where;
k₂/k₁ is the ratio of the rates which is the factor
Ea is the activation energy = 274 kJ/mol.
T₁ is the initial temperature = 231⁰C = 504 k
T₂ is the final temperature = 293⁰C = 566 k
R is gas constant = 8.314 J/Kmol
Substituting this values into the equation above;
ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)
ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)
ln(k₂/k₁) = 6.92
k₂/k₁ = exp(6.92)
k₂/k₁ = 1012.32
The rate of the reaction increased by 1012.32