Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
Answer:
V₂ = 2509.62 cm³
Explanation:
Given data:
Initial volume = 1500 cm³
Initial temperature = -65°C (-65 + 273 = 208 K)
Final temperature = 75°C ( 75 +273 = 348 K)
Final volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 1500 cm³ × 348 K / 208 k
V₂ = 522000 cm³.K / 208 k
V₂ = 2509.62 cm³
Answer:
Axis Labels
Explanation:
The axis labels are usually located on the x and y axis. This graph however is missing those.
hope this helps!
Answer:
Explanation:
Of course you could do the separation chemically. Dissolve the salt up in water, pass thru a filter, wash the iron filings with ethanol, which would encourage the salt to precipitate from solution.
I do hope I helped you! :)
