Answer:
pH = 4.8
Explanation:
A buffer is formed by a weak acid (0.145 M HC₂H₃O₂) and its conjugate base (0.202 M C₂H₃O₂⁻ coming from 0.202 M KC₂H₃O₂). The pH of a buffer system can be calculated using Henderson-Hasselbalch's equation.
![pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%5C%5CpH%20%3D%20-log%281.8%20%5Ctimes%2010%5E%7B-5%7D%20%29%2Blog%28%5Cfrac%7B0.202M%7D%7B0.145M%7D%20%29%5C%5CpH%3D4.8)
[A]0= Initial concentration
t1/2= half life
[A]= final concentration
k= rate constant
<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
I believe it's the second option. 2 or more elements joined together such that the elements have lost their individual identity in favour of a new set of properties.
If you start with 40.0 grams of the element at noon, 10.0 grams
radioactive element will be left at 2 p.m. The correct answer between
all the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
