5. A sinker made of 316 Stoinless Steel with an exact volume of 10.00mL has a ma
ID: 1074483 • Letter: 5
Question
5. A sinker made of 316 Stoinless Steel with an exact volume of 10.00mL has a mass of 79.900g in air. The sinker is suspended in a liquid sports drink that contains approximately 50% by weight protein. The mass of the sinker suspended in the liquid is only 67.500g. What is the density of the liquid protein drink? If the manufacturing release specification for density is 1.22-1.25, can the manufacturer release the product? If the praduct is to be sold in 1-pint sized serving containers, what is the weight of the liquid in each serving in both grams and pounds? The density of pure gold is 19.3g/cm2. What is the olume of 1 ounce of gold (A Troy ounce of gold is defined as 31.1 grams.) What is the mass of 1.00 oz of gold weighed in air. What is the mass of the same ounce of gold when suspended in water at 24.7 C? And.. in ethanol at the same temperature? (Refer to density tobles for water and ethonol.) 6. 7. The more exact formula for calculating the density of a Equid with a sinker is where is the weight correction factor(0.99985), to adjust for atmospheric buoyancy, and PL is the density of air (0.0012 g/cml. Why do you think these correction factors are not normally used in day-to-day testing in a manufacturing environment?Explanation / Answer
Archimedes principle will be used to calculate the density of liquid.
As per this principle, densty = mass of displaced liquid/volume of sticker
mass of 10ml sinker in air = 79.9 gm , mass of sinker suspended in liquid = 67.5 gm
change in mass of 10ml due to suspension in liquid = 79.9-67.5=12.4 gm
this is the mass of 10ml of liquid. Density of liquid = 12.4/10= 1.24 gm/ml
the specified density of manufacturer is in line with calculated density, the product can be released.
2. Density of gold= 19.3 g/cc, 1ounce of gold is= 31.1gm, volume of gold= mass/density = 31.1/19.3 cc=1.611399cc
1Oz= 29.5735ml, density of air at 24.7 deg,c can be calculated knowing density at STP = 29/22.4 =1.29 kg/m3 = 1.29*0.001 g/ml= 0.00129 g/ml, at 24.7 deg.c= 24.7+273= 297.7 K, density = 0.00129*273/297.7=0.001183 g/ml
mass of gold weighed in air = 29.5735ml*0.001183 g/ml=0.034985 gm
density of water= 1 g/ml, hence mass of gold suspended in water= 29.5735*1g/ml= 29.5735 gm
density of alcohol= 0.786g/ml, mass of gold suspended in alcohol= 29.5735*0.786 gm =23.244 gm
2. The desnity of air itself is very small and the correction factor that need to be applied is close to 1.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.