<h3>
Answer:</h3>
3.0 × 10²³ molecules AgNO₃
<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
- Writing Compounds
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
85 g AgNO₃ (silver nitrate)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃
The molecules are frozen in place but still vibrate
Zn⁰ ----> Zn⁺² + 2e⁻ - oxidation
Hg⁺² + 2e⁻ ----> Hg⁰ - reduction
Zn loses 2 moles of electrons , and Hg gains 2 mole of electrons.
So, number of moles of electrons gained and lost during reaction is equal.
Answer:
see explaination
Explanation:
Molecular equation;
2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)
Total ionic equation; . Includes all ions ;
6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)
Net ionic equation; remove common ions from total ionic;
2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)
Answer:- 3. 
and 
Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.
For example, the molecular formula of benzene is 
. The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.
In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is 
. In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.
In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is . The C to H ratio for second molecule is 1:4, so the empirical formula is 
. Here also, the empirical formulas are not same and hence it is also not the right choice.
In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also . Hence. this is the correct choice.
In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is . As the empirical formulas are different, it is not the right choice.
So, the only and only correct pair is the third one. 3. and
