An object accelerates from rest to a velocity of 34 m/s over a distance of 70 m. What was its acceleration?

6.8m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity  = 34m/s

Distance  = 70m

Unknown:

To solve this problem we apply one of the motion equations;

     V²   = U² + 2aS

where V is the final velocity

           U is the initial velocity

           a is the acceleration

         S is the distance covered

Input the parameters and solve;

   34² = 0² + 2 x a x 70

  34² = 140a

   a  = 6.8m/s²

b

explanation:

  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