Mole problems. See the image for reference. Please show your work.
1 answer:
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Explanation:
It is known that for , ppm present in 1
are as follows.
1 = 0.494 ppm
So, 150 =
= 0.15
Therefore, calculate the equivalent concentration in ppm as follows.
= 0.074 ppm
Thus, we can conclude that the equivalent concentration in ppm at STP is 0.074 ppm.
<h3>Answer:</h3>
13 g CO₂
<h3>General Formulas and Concepts:</h3><u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] 6.7 L O₂
[Solve] g O₂
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
[PT] Molar Mass of O: 16.00 g/mol
[PT] Molar Mass of C: 12.01 g/mol
Molar Mass of CO₂: 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
13.1637 g CO₂ ≈ 13 g CO₂