Answer:
The reaction is endothermic.
Yes, absorbed
3.06x10¹kJ are absorbed
Explanation:
In the reaction:
2HgO(s) → 2Hg(l) + O₂(g) ΔH = 182kJ
As ΔH >0,
<em>The reaction is endothermic</em>
<em />
As the reaction is endothermic, when the reaction occurs,
<em>the heat is absorbed.</em>
<em></em>
Now, based on the equation, when 2 moles of HgO (Molar mass: 216.59g/mol), 182kJ are absorbed.
72.8g are:
72.8g * (1mol / 216.59g) = 0.3361 moles HgO.
that absorb:
0.3361 moles HgO * (182kJ / 2 moles) =
<h3>3.06x10¹kJ are absorbed</h3>
The answer is (2) release a large amount of energy. Nuclear fission form light nuclides from heavy nuclides. While nuclear fusion form heavy nuclides from light nuclides.
MgCl2 would be magnesium chloride.
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>
