A particular brand of gasoline has a density of 0.737 g/mL at 25 ∘C. How many grams of this gasoline would fill a 15.7 gal tank
1 answer:
Th above problem asked us to find the amount of gasoline in grams given a density of 0.737 g/mL at 25°C and a volume of 15.7 gal.
To solve this problem, we must use the formula of density which is mass over volume.
D=
Base on the formula, we can find the mass of gasoline in by multiplying both sides by the volume. That is,
mass=D(Volume)
Note that we must cancel out the unit with a remaining g as the unit of mass. We must convert galloons to liters to mL.. That is
1 gal=3.78L and 1L=1000mL
15.7 gal x
x
=59346 mL
So,
mass=(0.737 g/mL x 59346 mL)=43738 g
To solve this problem, we must use the formula of density which is mass over volume.
D=
Base on the formula, we can find the mass of gasoline in by multiplying both sides by the volume. That is,
mass=D(Volume)
Note that we must cancel out the unit with a remaining g as the unit of mass. We must convert galloons to liters to mL.. That is
1 gal=3.78L and 1L=1000mL
15.7 gal x
So,
mass=(0.737 g/mL x 59346 mL)=43738 g
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Answer:
Approximately 6.81 × 10⁵ Pa.
Assumption: carbon dioxide behaves like an ideal gas.
Explanation:
Look up the relative atomic mass of carbon and oxygen on a modern periodic table:
- C: 12.011;
- O: 15.999.
Calculate the molar mass of carbon dioxide :
.
Find the number of moles of molecules in that sample of
:
.
If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:
,
where
is the pressure inside the container.
is the volume of the container.
is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
is the ideal gas constant.
is the absolute temperature of the gas.
Rearrange the equation to find an expression for , the pressure inside the container.
.
Look up the ideal gas constant in the appropriate units.
.
Evaluate the expression for :
.
Apply dimensional analysis to verify the unit of pressure.
Answer:
Trial Number of moles
1 0.001249mol
2 0.001232mol
3 0.001187 mol
Explanation:
To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.
Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.
<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.
In this case the solute is <em>NaOH</em>.
The formula is:
Solve for the <em>number of moles:</em>
Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.
Trial mL liters Number of moles
1 12.49 0.01249 0.01249liters × 0.1000M = 0.001249mol
2 12.32 0.01232 0.01232liters × 0.1000M = 0.001232mol
3 11.87 0.01187 0.01187liters × 0.1000M = 0.001187 mol