Answer:
The average force on ball by the golf club is 340 N.
Explanation:
Given that,
Mass of the golf ball, m = 0.03 kg
Initial speed of the ball, u = 0
Final speed of the ball, v = 34 m/s
Time of contact, 
We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

So, the average force on ball by the golf club is 340 N.
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
Answer: D) It should be handled in a fume hood, away from open flames.
Where are the statements?