<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
Where is the following statements??
The type of bonds present in the compound. and the type of structure it has and the elements that are presents and the number of moles of each element in one mole of the compound.
Answer:
23.8g
Explanation :
Convert 2.0M into mol using mol= concentration x volume
2.0M x 0.1L (convert 100mL to L since the units for M is mol/L)
= 0.2 mol
We can now find grams by using the molar mass of KBr
=119.023 g/mol (Found online) webqc.org
but can be be calculated by using the molecular weight of K and Br found on the periodic table
We can now calculate the grams by using grams=mol x molar mass
119.023g/mol x 0.2mol
= 23.8046 g
=23.8g (rounded to 1decimal place)
((|Vtrue-Vobserved|) / (Vtrue)) x 100
1.77777%error