Reduced- in a way losing an H means gaining an e-
NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
<span>Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% </span>
<span>3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
</span><span>0.8211 grams Na + 1.266 grams Cl = 2.087 grams</span>
<h3>
Answer:</h3>
2.04 mol CBr₄
<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>Organic</u>
- Writing Organic Compounds
- Writing Covalent Compounds
- Organic Prefixes
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
675 g CBr₄
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of Br - 79.90 g/mol
Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol
<u>Step 3: Convert</u>
<u /><u />
<u />
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄
Answer:
B. Cu (s) +Ni(NO3)2 (aq) - 2CuNO3 +Ni (s)
Explanation:
the above reaction is a substitution reaction