Answer:
Please find attached the completed Lewis dot diagram structure for PI₂F
Explanation:
The number of valence electrons are;
Phosphorus = 5 Electrons
Iodine = 2 × 7 electrons = 14 electrons
Chlorine = 7 electrons
The total number of valence electrons = 14 + 7 + 5 = 26 electrons
2) We draw the symbol that represents the basic (general) structure of the molecule as follows;
The sheared electron pair are represented by single bond lines
3) We complete the octet structures round the fluorine and the iodine atoms as attached showing 18 electrons plus 6 shared electron pairs, with a maximum from step 2 to give a total of (18 + 6) 24 electron pairs
4) We add the 2 unaccounted valence electron on the phosphorus atom to give it the stable octet structure, which gives the completed Lewis structure
Answer:
Latitude and longitude are imaginary lines that help us label every place on the surface of the earth. The most important line of latitude is the equator, which runs horizontally around the fattest part of the earth.
Explanation:
Explanation:
Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.
A. One of the oxides (Oxide 1) contains 63.2% of Mn.
Mass of the oxide = 100g
Mass of Mn = 63.2 g
Mass of O = 100 - 63.2
= 36.8 g
Ratio of Mn to O = 63.2/36.8
= 1.72
Another oxide (Oxide 2) contains 77.5% Mn.
Mass of oxide = 100 g
Mass of Mn = 77.5 g
Mass of O = 100 - 77.5
= 22.5 g
Ratio of Mn to O = 77.5/22.5
= 3.44
Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.
B.
Oxide 1
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Oxide 2
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Answer:
the spiral type of galaxy
Answer:
0.00335 moles
Explanation:
From the question, Using
PV = nRT................... Equation 1
Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³
Constant: R = 0.082 atm·dm³/K·mol
Substitute into equation 2
n = (1×0.075)/(273×0.082)
n = 0.075/22.386
n = 0.00335 moles
