<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399
<u>Explanation:</u>
Mass defect is defined as the difference in the mass of an isotope and its mass number.
The equation used to calculate mass defect follows:
![\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M](https://tex.z-dn.net/?f=%5CDelta%20m%3D%5B%28n_p%5Ctimes%20m_p%29%2B%28n_n%5Ctimes%20m_n%29%5D-M)
where,
= number of protons
= mass of one proton
= number of neutrons
= mass of one neutron
M = mass number of element
We are given:
An isotope of phosphorus which is 
Number of protons = atomic number = 15
Number of neutrons = Mass number - atomic number = 31 - 15 = 16
Mass of proton = 1.00728 amu
Mass of neutron = 1.00866 amu
Mass number of phosphorus = 30.973765 amu
Putting values in above equation, we get:
![\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399](https://tex.z-dn.net/?f=%5CDelta%20m%3D%5B%2815%5Ctimes%201.00728%29%2B%2816%5Ctimes%201.00866%29%5D-30.973765%5C%5C%5C%5C%5CDelta%20m%3D0.27399)
Hence, the mass defect for the formation of phosphorus-31 is 0.27399
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
The three isomers of pentane have different structural formulas.
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
