The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed
Answer:
19.8% of Nitrogen
Explanation:
In the Al(NO₃)₃ there are:
1 atom of Al
3 atoms of N
And 9 atoms of O
The molar mass of Al(NO₃)₃ is:
1 Al * (26.98g/mol) = 26.98g/mol
3 N * (14g/mol) = 42g/mol
9 O * (16g/mol) = 144g/mol
26.98 + 42 + 144 = 212.98g/mol
We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:
2.57g * (42g/mol / 212.98g/mol) =
0.51g N
Percent composition of nitrogen is:
0.51g N / 2.57g * 100
= 19.8% of Nitrogen
Identical electron configurations : K⁺ and Cl⁻
<h3>Further explanation </h3>
In an atom, there are levels of energy in the shell and sub-shell
This energy level is expressed in the form of electron configurations.
Charging electrons in the sub-shell uses the following sequence:
<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc. </em>
S²⁻ : [Ne] 3s²3p⁶
Cl : [Ne] 3s²3p⁵
K⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶
Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶
S :[Ne] 3s²3p⁴
Ar : [Ne] 3s²3p⁶
Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶
K : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
