Answer:
A - The can was open and gases were released.
KOH -------> K+ OH-
Ba(OH)2 ------> Ba+2. 2OH-
<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
Answer:
The answer is
<h2>112.912 kPa</h2>
Explanation:
The new pressure can be found by using the formula for Boyle's law which is
Since we are finding the new pressure
404.6 kPa = 404600 Pa
From the question we have
We have the final answer as
<h3>112.912 kPa</h3>
Hope this helps you
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the <u>cyclohexanone.</u>
See figure 1
I hope it helps!