An open train car, with a mass of 2050 2050 kg, coasts along a horizontal track at the speed 2.17 2.17 m/s. The car passes under a loading chute and, as it does so, gravel falls vertically into it for 2.85 2.85 s at the rate of 453 453 kg/s. What is the car's speed v f vf after the loading is completed? Ignore rolling friction.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

higher potential

explanation:

because it is in a high speed so it must move into a of high potential