An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are 355, 1480, and 5490 m/s, respectively. The sound travels a distance of 117 m in each medium. (a) After the first sound arrives, how much later does the second sound arrive

0.05773 sec

Explanation:

We know that

time = distance/speed = d/V

for air

t1 = 117/355 = 0.3295 sec

for fresh water

t_2 = 117/1480 = 0.07905 sec

for metal

t_3 = 117/5490 = 0.02131 sec

So as we can see sound will reach through metal hand rail 1st, then through fresh water and then through air

1st = metal hand rail

2nd = fresh water

3rd = air

time difference between 1st and 2nd sound = t_2 - t_3 = 0.07905 - 0.02131 = 0.05773 sec

i'm thinking more like energy

  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