So, it is seen that the destruction of ozone can be catalyzed by NO or by Cl. It has been found that the activation energy for t
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Answer:
Part 1
1. empirical formula is = N₂O₃
2. empirical formula is = NaClO₄
3. empirical formula is = BaCr₂O₇
Part2
no. of atoms of P₄ = 2.1 x 10²³
Part 3
A) no. of moles of S = 0.88 moles
B) no. of atoms of Mg = 1.08 x 10²⁴
C) no. of moles of Br₂ = 9.5 mole
Part 4
A) Molar mass of Na₂SO₄ = 142 g/mol
B) Molar mass of Al₂(SO₄)₃ = 342 g/mol
C) Molar mass of Al₂(SO₄)₃ = 176.5 g/mol
D) Molar mass of K₂CrO₄ = 194 g/mol
Part 5,
mass in grams of I₂ = 254 g
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Explanation:
Part 1:
Empirical Formula Calculation from %
1): Data Given
Percent mass of N = 63.6 %
Percent mass of O = 36.4 %
First convert percent to mass
let say we have 100 g of compound
So
mass of N = 63.6 /100 x 100 = 63.6 g
mass of O = 36.4 /100 x 100 = 36.4 g
Now convert masses to moles:
Molar mass of N = 14 g/mol
Molar mass of O = 16 g/mol
Formula used:
no. moles = mass in gram / molar mass ....................(1)
Now find the no. of moles of nitrogen
Put values in formula 1
no. moles = 36.4 g / 14 g/mol
no. moles = 2.6 mol
Now find the no. of moles of Oxygen
Put values in formula 1
no. moles = 63.6 g / 16 g/mol
no. moles = 4 mol
Now calculate the mole ratio of both element
N = 2.6 /2.6 = 1
O = 4 /2.6 = 1.5
To convert the ratio to whole number multiply the ratio with a whole number.
N = 1 x 2 = 2
O = 1.5 x 2 =3
So,
the ratio of N to O 2 : 3 and this is the simplest form
So the empirical formula is = N₂O₃
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2): Data Given
Percent mass of Na = 18.8 %
Percent mass of Cl = 29 %
Percent mass of O = 52.3 %
First convert percent to mass
let say we have 100 g of compound
So
mass of Na = 18.8 /100 x 100 = 18.8 g
mass of Cl = 29 /100 x 100 = 29 g
mass of O = 52.3 /100 x 100 = 52.3 g
Now convert masses to moles:
Molar mass of Na = 23 g/mol
Molar mass of Cl = 35.5 g/mol
Molar mass of O = 16 g/mol
Formula used:
no. moles = mass in gram / molar mass ....................(1)
Now find the no. of moles of Na
Put values in formula 1
no. moles = 18.8 g / 23 g/mol
no. moles = 0.82 mol
Now find the no. of moles of Cl
Put values in formula 1
no. moles = 29 g / 35.5 g/mol
no. moles = 0.82 mol
Now find the no. of moles of Oxygen
Put values in formula 1
no. moles = 52.3 g / 16 g/mol
no. moles = 3.3 mol
Calculate the mole ratio of both element
Na = 0.82 / 0.82 = 1
Cl = 0.82 / 0.82 = 1
O = 3.3 / 0.82 = 4
So,
The ratio of Na, Cl and O is 1 : 1 : 4 and this is the simplest form.
So the empirical formula is = NaClO₄
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3): Data Given
Percent mass of Ba = 38.9 %
Percent mass of Cr = 29.4 %
Percent mass of O = 31.7 %
First convert percent to mass
let say we have 100 g of compound
So
mass of Ba = 38.9 /100 x 100 = 38.9 g
mass of Cr = 29.4 /100 x 100 = 29 g
mass of O = 31.7 /100 x 100 = 31.7 g
Now convert masses to moles:
Molar mass of Ba = 137 g/mol
Molar mass of Cr = 52 g/mol
Molar mass of O = 16 g/mol
Formula used:
no. moles = mass in gram / molar mass ....................(1)
Now find the no. of moles of Ba
Put values in formula 1
no. moles = 38.9 g / 137 g/mol
no. moles = 0.28 mol
Now find the no. of moles of Cr
Put values in formula 1
no. moles = 29.4 g / 52 g/mol
no. moles = 0.56 mol
Now find the no. of moles of Oxygen
Put values in formula 1
no. moles = 31.7 g / 16 g/mol
no. moles = 2 mol
Calculate the mole ratio of both element
Ba = 0.28 / 0.28 = 1
Cr = 0.56 / 0.28 = 2
O = 2 / 0.28 = 7
So,
The ratio of Ba, Cr and O is 1 : 2 : 7 and this is the simplest form.
So the empirical formula is = BaCr₂O₇
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****Note: the rest of the answer is in attachment.
