P=0.0902 g/l
v=22.4 l/mol (stp)
M=vp
M=22.4 l/mol * 0.0902 g/l=2.020 g/mol
M=2.020 g/mol
<u>Answer 2 :</u> The given electronic configuration for a neutral atom of phosphorous in its ground state is incorrect.
Explanation :
A neutral atom of phosphorous has 15 electrons.
The given electronic configuration is incorrect.
The reason is, According to Aufbau principle, the electrons will be first filled in the sub-shell having lower orbital energy. As from the given configuration, 3p sub-shell has lower orbital energy than 4s sub-shell. So, the electrons will be filled in 3p sub-shell first. Hence, the ground state electronic configuration of neutral atom of phosphorous is,

<u>Answer 3 :</u>
Element Rubidium Magnesium Aluminium
Symbol Rb Mg Al
Group number 1 2 13
Number of valence 1 2 3
electrons
The order of general reactivity on the basis of number of valence electrons.
Rb > Mg > Al
Reason : The reactivity is determined by the number of electrons present in the outermost shell that means the element which have 1 valence electron will be more reactive because they can easily lose electrons.
Here, the three different notation of the p-orbital in different sub-level have to generate
The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number
depends upon the value of l, which are -l to +l.
Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as
,
and
in space.
The three p-orbital can be distinguish by the quantum numbers as-
For 2p orbitals (principal quantum number is 2)
1) n = 2, l = 1, m = -1
2) n = 2, l = 1, m = 0
3) n = 2, l = 1, m = +1
Thus the notation of different p-orbitals in the sub level are determined.
Answer:
A<em>=</em><em>4</em>
<em>B</em><em>=</em><em> </em><em>4</em><em>+</em><em>2</em><em> </em><em>=</em><em>6</em>
<em>C</em><em>=</em><em>C</em><em>arbon</em>
<em>D</em><em>=</em><em>10</em>
<em>E</em><em>=</em><em>22</em>
<em>F</em><em>=</em><em>Argon</em><em> </em>
Balanced chemical equation is
3CaCl2 +2Na3PO4-->6NaCl +Ca3(PO4)2
moles of CaCl2 =89.3g/[ (35.5x2) +40]=0.805moles
from the equation above the ratio of CaCl2 to Ca3(PO4)2 is 3:1 therefore the moles of Ca3(PO4)2 is 0.809/3=0.268moles
mass is therefore 0.268 x310.18(R.F.M of Ca3(PO4)2 ) =83.23grams