The answer is C which is PbSo
<h3>Answer:</h3>
7.57 × 10⁻²² g of F
<h3>Solution:</h3>
Data Given:
Number of Molecules = 8
M.Mass of BF₃ = 67.82 g.mol⁻¹
Mass of Fluorine atoms = ?
Step 1: Calculate Moles of BF₃
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Putting value,
Moles = 8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Moles = 1.33 × 10⁻²³ mol
Step 2: Calculate Mass of BF₃:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹
Mass = 9.0 × 10⁻²² g
Step 3: Calculate Mass of Fluorine Atoms:
As,
67.82 g BF₃ contains = 57 g of F
So,
9.0 × 10⁻²² g will contain = X g of F
Solving for X,
X = (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g
X = 7.57 × 10⁻²² g of F
In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,
Answer:
The percent yield of this reaction is 84.8 % (option A is correct)
Explanation:
Step 1: Data given
The student isolated 15.6 grams of the product = the actual yield
She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams
Step 2: Calculate the percent yield
Percent yield = (actual yield / theoretical yield ) * 100 %
Percent yield = (15.6 grams / 18.4 grams ) * 100 %
Percent yield = 84.8 %
The percent yield of this reaction is 84.8 % (option A is correct)
Answer:
29.42 Litres
Explanation:
The general/ideal gas equation is used to solve this question as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K
According to the information provided in this question;
mass of nitrogen gas (N2) = 25g
Pressure = 0.785 atm
Temperature = 315K
Volume = ?
To calculate the number of moles (n) of N2, we use:
mole = mass/molar mass
Molar mass of N2 = 14(2) = 28g/mol
mole = 25/28
mole = 0.893mol
Using PV = nRT
V = nRT/P
V = (0.893 × 0.0821 × 315) ÷ 0.785
V = 23.09 ÷ 0.785
V = 29.42 Litres
