The molar volume of any ideal gas at STP condition is 22.4 L
3.20 mol x 22.4L / 1 mol = 71.68 L = 71.7 L
1.59 moles
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<span>Hess' Law states that the enthalpy change in a reaction can be calculated from the enthalpy changes of reactions that, when combined, result in the desired reaction.
For example, to check the enthalpy change that occurs when benzene undergoes incomplete combustion to water and carbon monoxide is not an easy task, because the products invariably contain CO2. However, by combining the reactions of the complete combustion of benzene and the combustion of CO, you can get the reaction you want.
Reaction wanted: 2C6H6 + 9O2 → 12CO + 6H2O
Reactions provided: 2C6H6 + 15O2 → 12CO2 + 6H2O and 2CO + O2 → 2CO2, and their associated ΔH.
Rearrange the reactions so that, when they add up, they result in the wanted reaction.
2C6H6 + 15O2 → 12CO2 + 6H2O (leave as is; no changes to ΔH)
12CO2 → 12CO + 6O2 (reverse and multiply by 6; this changes the sign of ΔH and multiplies it by 6)
Added up, it will result in 2C6H6 + 9O2 → 12CO + 6H2O. Add up the ΔH values for the rearranged reactions to find ΔH for this particular reaction.</span>
<h3>
Answer:</h3>
2 M
<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>Unit 0</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
36.7 g CaF₂
300 mL H₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol
1000 mL = 1 L
<u>Step 3: Convert</u>
<em>Solute</em>
- Set up:

- Multiply:

<em>Solution</em>
- Set up:

- Multiply:

<u>Step 4: Find Molarity</u>
- Substitute [M]:

- Divide:

<u>Step 5: Check</u>
<em>Follow sig fig rules and round.</em> <em>We are given 1 sig fig as our lowest.</em>
1.56677 M ≈ 2 M
Answer:
because you'll be able to know if the reactants are oxidizing or reducing agents and you'll know if the products have been reduced or oxidized
Explanation:
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