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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<u>Answer:</u> The heat required for the process is 4.24 kJ
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of benzene = 24.8 g
Molar mass of benzene = 78.11 g/mol
Putting values in above equation, we get:
To calculate the enthalpy change of the reaction, we use the equation:
where,
= amount of heat absorbed = ?
n = number of moles = 0.318 moles
= enthalpy change of the reaction = 30.7 kJ/mol
Putting values in above equation, we get:
Hence, the heat required for the process is 4.24 kJ
Your answer is 3.25 moles of Bromine
