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
The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
<h3>Calculating mass </h3>
From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment
From the given information
Mass of empty evaporating dish = 46.233g
Mass of evaporating dish + Sodium bicarbonate = 48.230g
∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]
Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g
Mass of sodium bicarbonate (NaHCO₃) = 1.997 g
Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
Learn more on Calculating mass here: brainly.com/question/15268826
Solution here,
Volume(V)=67.4 L
Pressure(P)=1 atm
Temperature(T)=(0+273)K=273K
Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1
No. of moles(n)=?
Now,
PV=nRT
or, 1×67.4=n×0.0821×273
or, 67.4=22.4n
or, n=67.4/22.4
or, n=3
therefore, required no. of mole is 3.
V1 = 30 mL
P1 = 760 torr
P2 = 1520 torr
V2 = ?
applying Boyle's Law
P1*V1 = P2*V2
760 torr * 30 mL = 1520 torr * V2
V2 = 760 torr * 30 mL / 1520 torr
( C ) is correct
