A solution is prepared by mixing 30 ml pentane (C5H12, M. wt = 72 g/mol, density = 0.63g/mL) with 50 ml hexane as solvent (C6H14, M. Wt = 86 g/mol, density = 0.66g/cm, assuming that the volume added on mixing, calculate the mass percent of pentane?
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- At the surface of a freshwater lake, the pressure is 105 kPa What is the pressure increase in going 42.6 m below the surface? The density of fresh water at 4°C is 1.00 × 103 kg/m3. answer in kPaTwo large tanks, each holding 100L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3L/min and from B into A at a rate of 1L/min. The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.1kg/L of salt flows into tank A at a rate of 6L/min. The (diluted) solution flows out of the system from tank A at 4L/min and from tank B at 2L/min. If initially, tank A contains pure water and tank B contains 10kg of salt, determine the mass of salt in each tank at time t≥0. What is the solution to the system? x(t)= y(t)=What is maximum load a helium balloon can carry
- The density of aluminum is 2.70 g/cm^3. What is its aluminum's density in lb/yd^3?Water enters a mixing device at 150 L/s through pipe A, while oil of S.G. 0.8 is forced in at 30 L/s through a pipe B. If the liquids are incompressible and form a homogenous mixture, determine the average velocity and mass density of the mixture leaving through a pipe C of diameter 30 cm. Assume that there is no chemical reaction with the oil and water and that the mixture is incompressible.There is a liquid mixture at 20 ° C of 40% acetic acid and 60% by mass of water. Calculate the density of the mixture
- Which of the following is true for the situation below? po is the atmospheric pressure, p is the density of the mercury. Pgas - Mercury 16 cm 6 ст Pgas = PO Pgas + Pg(0.06m) = PO Pgas = Po + pg(0.10m) Pgas = Po + pg(0.16m) O Pgas + Pg(0.10m) = Po %3D pgas + pg(0.16m) = Po Pgas = Po + pg(0.06m)Problem 1: A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheric pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 2.15 × 10-4 m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.41 × 105 Pa.For this problem, assume that the density of air maintains a constant value of ρa = 1.20 kg / m3 and that the density of seawater maintains a constant value of ρs = 1025 kg / m3. What is the maximum depth dmax in meters below the surface of the ocean that the container can be taken before imploding?If the object has a mass of 1135 g and the volume of the water displaced is 250 cc, what is the object made of? material density (g/cc) Aluminum 2.70 Titanium 4.54 Nickel 8.90 Copper 8.96 Steel 7.87 Ausenitic stainless steel 7.8-8.0 Ferric stainless steel 7.7-7.8 Martensitic stainless steel 7.9-8.1 O Steel O Nickel O Titanium O Aluminum