A box at rest is in a state of equilibrium half way up on a ramp. The ramp has an incline of 42° . What is the force of static friction acting on the box if box has a gravitational force of 112.1 N ?

Explanation:

The force of static friction acting on the box is the frictional force;

Frictional force Ff = Wsin theta (force acting along the ramp)

W is the gravitational force known as the weight

Ff = 112.1sin42°

Ff = 112.1(0.6691)

Ff = 75.00N

Hence he force of static friction acting on the box  if box has a gravitational force of 112.1 N id 75.00N

  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

where is the