Answer:
sorry but i am doing this for point
Explanation:
how do I complete this column graph of the number of conduct vs the number of
<h3>
Answer:</h3>
0.6 g NaCl
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)
[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂
<u>Step 2: Identify Conversions</u>
[RxN] Na₂CO₃ → 2NaCl
Molar Mass of Na - 22.99 g/mol
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
0.551373 g NaCl ≈ 0.6 g NaCl
i think that it is the tree
4.0
i think it has something to do with molar ratios and finding the limiting reactant
4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2
4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2
so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO
once the limiting reactant is found, we can use that data for that substance to calculate the amount of product
4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2
mass = mol no. x molar mass
= 13.6 x 56+(3x14)+(9x16)
= 242g