This question deals with waste disposal in the Solutions and Spectroscopy experiment. What should be done to waste solutions con
1 answer:
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Answer:
<em>Option C: 4.54 x </em><em> KJ/mol of nuclei</em>
<em>Note: </em>Here in this question option C is not correctly put. It is 4.54 x rather than 4.54 x
.
Explanation:
If mass defect is known, then nuclear binding energy can easily be calculated, here's how:
First step is to convert that mass defect into kg.
Mass defect = 5.0446 amu
Mass defect = 5.0466 x 1.6606 x
Because 1 amu = 1.6606 x Kg.
<em>Mass defect = 8.383 x </em><em> kg.</em>
Now, we need to find out it's energy equivalent by using following equation:
Using the equation E = mc²:
where c= 3.00 x m/s²
E = (8.383 x ) x (3.00 x
)²
E = 7.54 x J this energy is in Joules but nuclear binding energy is usually expressed in KJ/mol of nuclei. Let's convert it:
(7.54 x Joule/nucleus)x(1 kJ/1000 Joule)x(6.022 x
nuclei/mol) =
<em>4.54 x </em><em> kJ/mol of nuclei .</em>
E = <em>4.54 x </em><em> kJ/mol of nuclei .</em> So, this is the nuclear binding energy of that atom, which is option C.
<em>Note:</em> Here in this question option C is not correctly put. It is 4.54 x rather than 4.54 x
Answer: Mass of required to form 930 kg of iron is 1328 kg
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
For iron:
Given mass of iron = 930 kg = 930000 g (1kg=1000g)
Molar mass of iron = 56 g/mol
Putting values in equation 1, we get:
The chemical equation for the production of iron follows:
By Stoichiometry of the reaction:
2 moles of iron are produced by = 1 mole of
So, 16607 moles of iron will be produced by = of
Now, calculating the mass of from equation 1, we get:
Mass of =
Thus mass of required to form 930 kg of iron is 1328 kg