Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
<span>The concentration of pb2+ = 1.00mg/ml
Diluted Solution is 6.0 x 102 ml = 612 ml
Volume of the concentration of pb2+ is 0.054 mg/l is v
(vL)(1.00mg/ml) = (.612L)(0.054mg/l)
Volume = 0.033048L
Volume of the concentration of pb2+ is 0.054 mg/l = 33.048 ml.</span>
Answer:
Mole Fraction (H₂O) = 0.6303
Mole Fraction (C₂H₅OH) = 0.3697
Explanation:
(Step 1)
Calculate the mole value of each substance using their molar masses.
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g
Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol
Molar Mass (C₂H₅OH): 46.068 g/mol
300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g
(Step 2)
Using the mole fraction ratio, calculate the mole fraction of each substance.
moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent
11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (H₂O) = 0.6303
6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (C₂H₅OH) = 0.3697
