The two compounds shown indeed have tha same molecular formula, C5 H11 NO2. One of the molecules has a group NH2 and a group COOH, the other molecule has a NOO group, that makes that the two isomers have a completely different structure, with the atoms arranged in a completely different order. <span>This kind of isomers fits in the definition of structural isomers, so the answer is structural isomers.</span>
Answer:
Explanation:
In this question, we wish to find the missing nuclei for the equation:
In order to find the missing species, we need to use the charge and mass balance law. That is, the mass should be conserved: the total mass on the left-hand side with respect to the arrow should be equal to the total mass on the right-hand side with respect to the arrow:
Notice from here that:
So far we know that the mass of X is 4. Similarly, we apply the law of charge conservation. The total charge should be conserved:
From here:
We have a particle:
Looking at the periodic table, an atom with Z = 2 corresponds to helium. This can also be written as an alpha particle:
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
