A car is on an inclined plane at an angle of 15 degrees. The car has a free body diagram with three vectors coming from it. The first is parallel to the incline and up the incline labeled F Subscript f Baseline. The force upward and perpendicular to the car is labeled F Subscript N Baseline. The third force pointed down is labeled F Subscript g Baseline and is broken up into two perpendicular components labeled F Subscript g y Baseline and the other F Subscript g x Baseline. Examine the forces acting on a car on a ramp. Select the forces or force components that need to be added together to calculate the net force in the x-direction. Check all that apply. FN
–Ff
–Fg
Fg (cos 15°)
Fg (sin 1

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz