<h3>
Answer:</h3>
0.0157 g Au
<h3>
General Formulas and Concepts:</h3>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.113 g Au
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 197.87 g/mol
<u>Step 3: Convert</u>
<u />
= 0.015733 g Au
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.015733 g Au ≈ 0.0157 g Au
Answer:
f = 33.34 Hz
Explanation:
A wave has a period of 0.03 seconds. It is required to find the frequency of a wave. The relation between time period and frequency is inverse. The time period of a wave is given by :
T = 1/f, f = frequency of wave
So, the frequency of the wave is 33.34 Hz.
Answer:
15.4%
Explanation:
If Ka = 0.54 mM = 1.51x10⁻⁵
Then;
C₄H₈O₂ --------> C₄H₇O₂⁻ + H⁺
I 0.54x10⁻³ 0 0
E 0.54x10⁻³(1-x) 0.54x10⁻³x 0.54x10⁻³x
Recall that x is the percentage degree of dissociation
From the ICE table;
Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]
1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)
1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x
1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2
1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2
Hence;
0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0
x^2 + 0.028x - 0.028 = 0
Solving the quadratic equation here;
x = 0.154 or −0.182
Ignoring the negative result, x = 0.154
Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%
Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.
