You and your friend are running at each other from 200m apart. after 10 seconds and having passed the 50m mark on your side, you see your friend has already gone 60m on his. what will your average velocity need to be for the remaining time in order for you to both have covered the same amount of distance, assuming he maintains a constant speed?

average speed = 7.50 m/s

Explanation:

given data

distance = 200 m

time = 10 seconds

passed = 50m

already gone = 60 m

to find out

average velocity need to be for the remaining time

solution

we get here first Velocity of friend v that is express as

velocity =

velocity =

velocity = 6 m/s

and time taken to reach half the way is here as

time taken =

time taken = 16.67 seconds

and now distance to be covered is

distance to be covered =  100 m - 50 m

distance to be covered = 50 meters

and remaining distance cover in time = 16.67 sec - 10 sec

remaining distance cover in time = 6.67 seconds

so average speed will be here as

average speed =  

average speed = 7.50 m/s

  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