Less, because thermal energy is heat so if it gets colder there is less thermal energy.
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:
12*2*6.022*10^23. Atoms of hydrogen.
Explanation:
First of all if u know that 1 molecule of calcium hydroxide contain 2 hydrogen atom and 1 mole of calcium hydroxide contain 6.022*10^23 molecules of calcium hydroxide.
Hence 1 mole of calcium hydroxide contain 2*6.022*10^23 atom of hydrogen.
So 12 mole of calcium hydroxide contain =12*2*6.022*10^23. Atoms of hydrogen.
Answer:
tin (IV) oxide octahydrate contains 8 part water
What is the percent composition of water found in tin (IV) oxide octahydrate?
