A mass is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the mass and let go. It takes the mass 0.50 s to go back to its equilibrium position when it is oscillating. What is the frequency of oscillation?

0.5 Hz

Explanation:

If it takes 0.5s to go back to the equilibrium position, that means it would take 1 s to reach maximum compression, and 2s to go back to the initial stretching position, aka complete 1 cycle of oscillation.

So the period is T = 2s. Then the frequency would be

  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

the answer would be a