1. Rolls of aluminum foil are 301 mm wide and 0.017 mm thick. What maximum lengt
ID: 1001466 • Letter: 1
Question
1. Rolls of aluminum foil are 301 mm wide and 0.017 mm thick. What maximum length of aluminum foil can be made from 0.94 kg of aluminum?( d(aluminum)=2.70 g/cm3) Express your answer using two significant figures.
2. Lead metal can be extracted from a mineral called galena, which contains 86.6% lead by mass. A particular ore contains 68.5%galena by mass.If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.00 cm radius?
Express your answer with the appropriate units.
3. Calculate the mass, in kg, of each sample. 1.93×1027 nickel atoms.
Express your answer with the appropriate units.
4. What is the mass, in grams, of each of the following? 43.9 mol Xe
Express your answer using three significant figures.
5. What is the mass, in grams, of each of the following? 1.1 mol W
Express your answer using two significant figures.
6. 5.98×104 J to Cal
Express the value in Calories to three significant figures.
Explanation / Answer
Rolls of aluminum foil are 301 mm wide and 0.017 mm thick. What maximum length of aluminum foil can be made from 0.94 kg of aluminum?( d(aluminum)=2.70 g/cm3) Express your answer using two significant figures.
Density = mass / volume
volume = mass/density
volume = length x width x height
here density = 2.70 g/cm3
width = 301 mm = 30.1 cm
thick = 0.017 mm= 0.0017 cm
length = ?
first calculate the volume = mass/density
= 0.94 *10^3 g/ 2.70 g/cm^3
= 348.15 cm^3
length = volume / width x height
=348.15 cm^3 / 30.1 cm *0.0017 cm
= 6803.7 cm
= 68.04 m
2. Lead metal can be extracted from a mineral called galena, which contains 86.6% lead by mass. A particular ore contains 68.5%galena by mass.If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.00 cm radius?
Express your answer with the appropriate units.
Density of Lead is 11.34 grans / cm^3
Density = mass / volume
Mass = density * volume
Volume = 4/3 pi r ^3
= (4/3) * pi * (3.00 cm)^3 * [ 11.34 gm / cm^3 ] * [ 100/86.6 ]*[ 100 / 68.5] * [ 100 / 92.5]
here pi = 3.14
= 2336.12 g
=2.34 kg
3. Calculate the mass, in kg, of each sample. 1.93×1027 nickel atoms.
Express your answer with the appropriate units.
1.93×1027 nickel atoms.
Molar mass of Ni = 58.69 g/ mol
First calculate the number of moles
= 1.93*10^27 atoms * 1 mole/ 6.023*10^23atoms
= 3097.91 moles
Amount of 3097.91 moles Ni
3097.91 moles *58.69 g/ mol
=181816.5 g
= 181.82 kg
4. What is the mass, in grams, of each of the following? 43.9 mol Xe
Express your answer using three significant figures.
Molar mass of Xe = 131.293 g/ mol
the mass, in grams, 43.9 mol Xe = 131.293 g/ mol * 43.9 mol Xe
= 5763.8 g
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