The answer is B plastic bags made from petroleum.
Heat
gained in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. It is
expressed as follows:<span>
Heat = mC(T2-T1)
When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate things. We do as follows:
</span>Heat gained = Heat lost
mC(T2-T1) = - mC(T2-T1)
31.5C (102.4 - 32.5) = 103.5(4.18)(32.5 - 24.5)
C = 1.57 J/C-g
Hope this helps.
The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
It is non-polar molecule.
CF₄ - dipole moment = 0D
Answer : The percentage reduction in intensity is 79.80 %
Explanation :
Using Beer-Lambert's law :



where,
A = absorbance of solution
C = concentration of solution = 
l = path length = 2.5 mm = 0.25 cm
= incident light
= transmitted light
= molar absorptivity coefficient = 
Now put all the given values in the above formula, we get:



If we consider
= 100
then, 
Here 'I' intensity of transmitted light = 20.198
Thus, the intensity of absorbed light
= 100 - 20.198 = 79.80
Now we have to calculate the percentage reduction in intensity.


Therefore, the percentage reduction in intensity is 79.80 %