How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP
2 answers:
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in of ethane gas
And, as we know that
1 mole of ethane molecule contains molecules of ethane
2.869 moles of ethane molecule contains molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
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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