CHM 1045 Lab Exam 1 ACTUAL UPDATED QUESTIONS AND CORRECT ANSWERS
Exp 2- Density Equation D=m/v
Exp 2- Vcylinder Equation Vcylinder=pi r^2 h
Exp 2- Vsphere Equation Vsphere= 4/3 pi r^3
Exp 2- What does r, h, and pi stand for? R= radius
H= height
pi= 3.142....
Exp 2- What is cm^3 equal to? mL
Exp 2- Equation for Mass of Water Mass of water= (mass of beaker + water) - (mass of beaker)
Exp 2- Equation for Mass of Unknown Liquid Mass of unknown liquid= (mass of beaker + unknown liquid) - (mass of beaker)
, Exp 2- What is the volume of the cylinder by water Volume of cylinder= (volume of water in graduated cylinder with unknown) -
displacement formula? (volume of original water in graduated cylinder)
Exp 2- If you have the diameter of the cylinder, what is Convert diameter to radius
your next step?
Exp 2- Equation for Radius of cylinder? r= d/2
Exp 2- If the measured volume of the cylinder was lower Higher than the true value
than the true volume, would the calculated density be (Explanation: d=m/v: if v is too low; d will be too high)
higher or lower than the true value?
Exp 2- What is the density in g/mL of a sphere which 1.) 2.00 lbs * 454 g/ 1 lb= 908 g
weighs 2.00 lb and has a diameter of 7.00 inches? 2.) 7.00 in= diameter (7.00 in/2= 3.50 in = radius)
3.) V= 4/3 pi r^3 ----- 4/3 pi (3.50)^3
4.) convert 3.50 in to cm
5.) 3.50 in * 2.54 cm/ 1 in= 8.89 cm
6.) Vcylinder= 4/3 pi (8.89cm)^3
7.) D=m/v--- 908g/2.94X 10^3 mL= .309 g/mL
Exp 2- Consider a stone column 3.00 ft in diameter and 1.) 3.00 ft diameter
10.00 ft high. What is its mass in pounds if the density is 2.) convert 3.00 ft to inches to cm---3.00 ft 12 in/1 ft 2.54 cm/1 in= 91.4 cm
3.50 g/mL? (diameter)
3.) 10.00 ft diameter
4.) convert 10.00 ft to inches to cm---10.00 ft 12 in/1 ft 2.54 cl/ 1 in= 304.8 cm
5.) V= pi (91.4 cm)^2 * (304.8 cm)
- ---------------------------------- = 2.00 x 10^6 cm^3
4
6.) 2.00 x 10^6 mL= 3.50g/1 mL x 1 lb/ 454g = 1.54 x 10^4 lbs
Exp 2- Suppose that a student performed the experiment The calculated value for the density of the unknown would not be correct. The
and the calculations perfectly as directed except that, balance needs to be zeroed out prior to each measurement. When you weigh the
unknown to the student, the balance was not zeroed but difference for the volume, the amount would cancel out.
weighed 0.100 g too high throughout the experiment.
Would the calculated value for the density of the
unknown liquid be correct? That is, would it be equal to,
higher than, or lower than the true value.
Exp 2- If the same balance that was used in question 4 The mass of the metal cylinder would be too high. Since the balance was not
was used to determine the mass of the metal cylinder, zeroed out when you place the cylinder on the balance the mass would be too
would the calculated value be equal to, higher than, or high.
lower than the true value?
Exp 2- Density Equation D=m/v
Exp 2- Vcylinder Equation Vcylinder=pi r^2 h
Exp 2- Vsphere Equation Vsphere= 4/3 pi r^3
Exp 2- What does r, h, and pi stand for? R= radius
H= height
pi= 3.142....
Exp 2- What is cm^3 equal to? mL
Exp 2- Equation for Mass of Water Mass of water= (mass of beaker + water) - (mass of beaker)
Exp 2- Equation for Mass of Unknown Liquid Mass of unknown liquid= (mass of beaker + unknown liquid) - (mass of beaker)
, Exp 2- What is the volume of the cylinder by water Volume of cylinder= (volume of water in graduated cylinder with unknown) -
displacement formula? (volume of original water in graduated cylinder)
Exp 2- If you have the diameter of the cylinder, what is Convert diameter to radius
your next step?
Exp 2- Equation for Radius of cylinder? r= d/2
Exp 2- If the measured volume of the cylinder was lower Higher than the true value
than the true volume, would the calculated density be (Explanation: d=m/v: if v is too low; d will be too high)
higher or lower than the true value?
Exp 2- What is the density in g/mL of a sphere which 1.) 2.00 lbs * 454 g/ 1 lb= 908 g
weighs 2.00 lb and has a diameter of 7.00 inches? 2.) 7.00 in= diameter (7.00 in/2= 3.50 in = radius)
3.) V= 4/3 pi r^3 ----- 4/3 pi (3.50)^3
4.) convert 3.50 in to cm
5.) 3.50 in * 2.54 cm/ 1 in= 8.89 cm
6.) Vcylinder= 4/3 pi (8.89cm)^3
7.) D=m/v--- 908g/2.94X 10^3 mL= .309 g/mL
Exp 2- Consider a stone column 3.00 ft in diameter and 1.) 3.00 ft diameter
10.00 ft high. What is its mass in pounds if the density is 2.) convert 3.00 ft to inches to cm---3.00 ft 12 in/1 ft 2.54 cm/1 in= 91.4 cm
3.50 g/mL? (diameter)
3.) 10.00 ft diameter
4.) convert 10.00 ft to inches to cm---10.00 ft 12 in/1 ft 2.54 cl/ 1 in= 304.8 cm
5.) V= pi (91.4 cm)^2 * (304.8 cm)
- ---------------------------------- = 2.00 x 10^6 cm^3
4
6.) 2.00 x 10^6 mL= 3.50g/1 mL x 1 lb/ 454g = 1.54 x 10^4 lbs
Exp 2- Suppose that a student performed the experiment The calculated value for the density of the unknown would not be correct. The
and the calculations perfectly as directed except that, balance needs to be zeroed out prior to each measurement. When you weigh the
unknown to the student, the balance was not zeroed but difference for the volume, the amount would cancel out.
weighed 0.100 g too high throughout the experiment.
Would the calculated value for the density of the
unknown liquid be correct? That is, would it be equal to,
higher than, or lower than the true value.
Exp 2- If the same balance that was used in question 4 The mass of the metal cylinder would be too high. Since the balance was not
was used to determine the mass of the metal cylinder, zeroed out when you place the cylinder on the balance the mass would be too
would the calculated value be equal to, higher than, or high.
lower than the true value?
