Jamie is hiking up a small mountain. He climbs up at a constant rate of 300 feet/hour until he reaches the peak at 1,500 feet. After that, he hikes down at the same rate to the base of the mountain. The equation that models Jamie’s elevation, e, after t hours is e =
a. - | t-300 | + 1500
b. -1500 | t - 5 | + 300
c. - | t - 1500 | + 300
d. - 300 | t - 5 | + 1500
. Jamie’s elevation will be 600 feet after
a. 2 b. 3 c. 4 d. 5
hours and after
a. 6 b. 7 c. 8 d. 9 hours.

  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