Object 1 is moving along the x axis with an initial momentum +16 kg·m/s, where the + sign indicates that it is moving to the right. as the drawing shows, object 1 collides with a second object that is initially at rest. the collision is not head-on, so the objects move off in different directions after the collision. the net external force acting on the two-object system is zero. after the collision, object 1 has a momentum whose x component is +6 kg·m/s. what is the x-component of the momentum of object 2 after the collision?

  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

f is prop to m1xm2 double m1 double force

your answer is, orbit around the nucleus