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<h3>Answer:</h3>
0.0253 mol H₂O
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[Given] 0.456 g H₂O (water)
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.025305 mol H₂O ≈ 0.0253 mol H₂O
Answer : The volume of produced at standard conditions of temperature and pressure is 0.2422 L
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of
gas = (740-22.4) torr = 717.6 torr
= final pressure of
gas at STP= 760 torr
= initial volume of
gas = 280 mL
= final volume of
gas at STP = ?
= initial temperature of
gas =
= final temperature of
gas =
Now put all the given values in the above equation, we get:
Therefore, the volume of produced at standard conditions of temperature and pressure is 0.2422 L