Answer:
inonic bonds with cavalent bonds
Explanation:
ionic bonds
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>
Answer:
the rate of reaction would decrease
Explanation:
The answer is eight electrons.
Option C: The dissociation of a polar covalent compound in water.
Sugar is made up of sucrose molecules containing polar covalent bonds. They have same type of oxygen-hydrogen covalent bonds that are present in water.
Since, sugar is not an ionic compound, option a and option b are incorrect. Also, due to same type of bonding in water and sugar molecules there will be no such force of attraction between them, this will opt out the option d.
Thus, dissolution of sugar in water is the dissociation of a polar covalent compound in water and option c is correct.
