<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>
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<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
I think the answer is A but I could be wrong
Answer:
% comp. H = 2.8%, % comp. Cl = 97.2%
Explanation:
HCl mass = mass of H + mass of Cl
HCL mass = 1.00794 + 35.453 = 36.46094
% comp. of H = 1.007694 / 36.46094 x 100 = around 2.8% (2.76376308455)
% comp. of Cl = 35.453 / 36.46094 x 100 = around 97.2% (97.2355622208)
Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>