And Answer what answer would be 3.14
Activation energy
Hope this helps
Answer: 136 g of
must be reacted to inflate an air bag to 70.6 L at STP.
Explanation:
Using ideal gas equation:
P= pressure of nitrogen gas = 1 atm (at STP)
V =volume of nitrogen gas = 70.6 L
n = number of moles of nitrogen gas = 1 atm (at STP)
R = gas constant = 0.0821 Latm/Kmol
T = temperature of nitrogen gas = 273 K (at STP)
For the balanced chemical reaction:

3 moles of nitrogen are produced by = 2 moles of 
3.14 moles of nitrogen are produced by =
moles of 
Mass of
= Moles × Molar mass = 2.09 mole × 65 g/mol = 136 g
The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.