The number of years required for 1/4 cobalt-60 to remain after decay is calculated as follows
after one half life 1/2 of the original mass isotope remains
after another half life 1/4 mass of original mass remains
therefore if one half life is 5.3 years then the years required
= 2 x 5.3years = 10.6 years
Well for a start, this makes absolutely no sense, "discovered a fuel that burns so hot that it becomes cold."
<span>And yes, it's not science if the experiment can't be repeated. In fact they should WANT it to be repeated so that you can get credit for discovering something new and then possibly harness this effect to produce useful applications. </span>
<span>For all we know they had a fewer of LN2 in the lab that got shredded by the blast, LN2 could certainly have frozen many things (not metal though, since metal is already solid at room temperature, (except for mercury)), and afterwards would leave no trace.</span>
Answer:
Option C. Energy Profile D
Explanation:
Data obtained from the question include:
Enthalpy change ΔH = 89.4 KJ/mol.
Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:
Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.
If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.
Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.
Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.