A zebra starts from rest and accelerates at 1.9 m/s2 what will be the acceleration after 5 seconds

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

Corrected question;

A zebra starts from rest and accelerates at 1.9 m/s2. How far has the zebra gone after 5 seconds.

Explanation:

From the equation of motion;

d = vt + 0.5at^2 1

Where;

d = displacement

v = initial velocity = 0

t = time taken= 5 seconds

a = acceleration = 1.9 m/s^2

Substituting the given values into equation 1;

d = 0(5) + 0.5(1.9×5^2)

d = 0.5(1.9×5^2)

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

  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

less than one you can't create energy