Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ
where, t1/2 is half-life and λ is the decay constant.
t1/2 = 10 min = 0.693 / λ
Hence, λ = 0.693 / 10 min - (1)
Nt = Nο e∧(-λt)
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken
by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)
From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500
Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%
Hence, Percentage of remaining nuclei is 25.00%
Answer:
Convert the mass of each element to moles using the molar mass from the periodic table. then Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number
Answer: option D - The total number of nucleons changes.
Explanation:
Nuclear Reaction is best described as a process such as the fission of an atomic nucleus, or the fusion of one or more atomic nuclei and / or subatomic particles in which the NUMBER of PROTONS and / or NEUTRONS in a nucleus CHANGES; the reaction products may contain a different element or a different isotope of the same element.
Note that the NUCLEONS refers to ONE of the subatomic particles of the atomic nucleus, i.e. a PROTON or a NEUTRON.
So, in a Nuclear reaction, the total number of nucleons changes.
Answer:
e
Explanation:
<em>Provided the reaction that leads to the formation of the products can proceed in both forward and backward directions, the correct answer would be yes because the reaction will proceed backward until equilibrium is reached.</em>
<u>For a reaction that can proceed both forward and backward, the addition of a catalyst increases the rate of reaction in both directions based on the fact that a catalyst cannot alter the equilibrium of a reaction. </u>
Hence, if an enzyme is added to the product of a reaction that has the potential to proceed in both forward and reverse reactions, a substrate would be expected to form because the reaction will proceed backward until an equilibrium is reached.
The correct option is e.
