Answer:
Isotope N–14 = 99%
Isotope N–15 = 1%
Explanation:
Let isotope A be N-14
Let isotope B be N-15
From the question given above, the following data were obtained:
For isotope A (N-14):
Mass of A = 14
Abundance of A = A%
For isotope B (N-15):
Mass of B = 15
Abundance of B = (100 – A%)%
Atomic mass of nitrogen = 14.01 amu
Thus, we can obtain the relative abundances of the naturally occurring isotopes of nitrogen as illustrated below:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
14.01 = [(14 × A%)/100] + [(15 × (100 – A%)/100]
14.01 = 0.14A% + 0.15(100 – A%)
14.01 = 0.14A% + 15 – 0.15A%
Collect like terms
14.01 – 15 = 0.14A% – 0.15A%
– 0.99 = – 0.01A%
Divide both side by – 0.01
A% = – 0.99 / –0.01
A% = 99%
Abundance of B = (100 – A%)%
Abundance of B = (100 – 99)%
Abundance of B = 1%
Thus, the relative abundances of the naturally occurring isotopes of nitrogen are:
Isotope N–14 = 99%
Isotope N–15 = 1%
Answer:
High melting point,High density,Nonreactive
Explanation:
The following are the Properties of transition elements:
They have large charge/radius ratio. They are harder and they have high densities;.They have high melting and boiling points.They form compounds which are paramagnetic. They show variable oxidation states. They form colored ions and compounds. Compounds are formed with profound catalytic activity. Stable complexes are formed by them.
When Comparing with the alkali metals present in group 1 and the alkaline Earth metals that are present in group 2, the transition metals are considered to be much less reactive. They do not react fast with water or oxygen, which contributes to their resistance to corrosion.
The amount of the precipitate PbCl2 can be obtained using stoichiometry, assuming the reaction goes into completion given the excess amounts of the lead (II) nitrate solution. First, divide 2.5 g NaCl to its MW of 58.44 g/ mol to obtain the moles of NaCl involved in the reaction. Second, knowing that for every 2 moles of NaCl, there is 1 mole of PbCl2 produced, we divide the moles of NaCl obtained earlier by 2 to get the moles of PbCl2 produced. From the moles of PbCl2, we multiply it to its MW of 278.1 g/ mol. The amount of precipitate is then calculated to be 5.9484 g PbCl2.