<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
There is 6 non - bonding pairs.
Let me show you one easy method to do this.
o22-, oxygen valence electron = 6 here we have two so total 12, and -2 that means we add electrons so it’s all equal to 14 right.
whenever need to find lone pair, subtract the number you get with the lowest multiple of 8.
here we total 14 valence electron right so lowest multiple of 8 would be 8.
so 14 - 8 = 6 and that is our answer.
Let me know if you have Problem with chemistry.
Answer: The osmotic pressure of a solution is 53.05 atm
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

Or,

where,
= osmotic pressure of the solution = ?
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (methanol) = 22.3 g
Volume of solution = 321 mL
R = Gas constant = 
T = temperature of the solution = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
Putting values in above equation, we get:


Hence, the osmotic pressure of a solution is 53.05 atm
B. False
Stability is determined by the ratio of neutrons and protons. Electrons are not in nucleus.
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:

Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g


The percent yield of chloro-ethane in the reaction is 82.98%.