Answer: 2 lone pairs, square planar
Explanation:
Using the VSEPR ( Valence Shell Electron Pair Repulsion)Theory
To calculate the number of lone pairs electron can be done using the formula;
Number of electrons = ½ (V+N-C+A)
V mean valency of the central atom
N means number of monovalent bonding atoms
C means charge on cation
A means charges on anion
Therefore, to calculate the number of lone pair electron C=A=0;
Number of electrons = ½ (8+4) = 12/2 = 6
Number of bonding pair = 4
Number of lone pairs of electron = 6-4 = 2
The hybridrization of the compound is sp3d2 because the number of electrons around the central atom is 6.
The geometry of the compound is square planar and this is because of the repulsion between the bonding pair of electrons and lone pair of electrons which causes the lone pair of electrons to lie in a perpendicular plane in order to acquire stability.
The correct response is C. Chlorine forms 1 covalent bond while Oxygen forms 2 covalent bonds.
Answer:
17.65 grams of O2 are needed for a complete reaction.
Explanation:
You know the reaction:
4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O
First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
So, the molar mass of the compounds in the reaction is:
- NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- NO: 14 g/mol + 16 g/mol= 30 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
By stoichiometry, they react and occur in moles:
- NH₃: 4 moles
- O₂: 5 moles
- NO: 4 moles
- H₂O: 6 moles
Then in mass, by stoichiomatry they react and occur:
- NH₃: 4 moles*17 g/mol= 68 g
- O₂: 5 moles*32 g/mol= 160 g
- NO: 4 moles*30 g/mol= 120 g
- H₂O: 6 moles*18 g/mol= 108 g
Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?
mass of O₂≅17.65 g
<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>
To solve this question,
let us first calculate how much all the nucleons will weigh when they are apart,
that is:
<span>Mass of 25 protons = 25(1.0073) = 25.1825 amu </span>
Mass of neutrons = (55-25)(1.0087) = 30.261 amu
So, total mass of nucleons = 30.261+25.1825 =
55.4435 amu
<span>Now we subtract the mass of nucleons and mass of the Mn
nucleus:
55.4435 - 54.938 = 0.5055 amu
This difference in mass is what we call as the mass defect of
a nucleus. Now we calculate the binding energy using the formula:</span>
<span> E=mc^2 </span>
<span>But first convert mass defect in units of SI (kg):
Δm = 0.5055 amu = (0.5055) / (6.022x10^26)
<span>Δm = 8.3942x10^-28 kg</span>
Now applying the formula,
E=Δm c^2
E=(8.3942x10^-28)(3x10^8)^2
E=7.55x10^-11 J</span>
Convert energy from Joules
to mev then divide by total number of nucleons (55):
E = 7.55x10^-11 J *
(6.242x10^12 mev / 1 J) / 55 nucleons
<span>E = 8.57 mev / nucleon</span>
