In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. a second projectile (of the same mass) causes the the pendulum to swing twice as high, h2 = 5.2 cm.
the second projectile was how many times faster than the first?

  Second  projectile is 1.4 times faster than first projectile.

Explanation:

By linear momentum conservation

Pi = Pf

m x U + M x 0 = (m + M) x V

Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at reference point

KEf = 0, Speed of system zero at height 'h'

PEf = (m + M) g h

So,

So from above value of V

Initial velocity of projectile =U

Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.

Which means

U₂ = 1.41 U₁

Therefore we can say that ,Second  projectile is 1.4 times faster than first projectile.

  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