The mass of mars is 6.38x10^23 kg and its radius is 3.38 x10^6m. mars rotates on its axis with a period of 1.026 days.(g=6.67x10^-11 nm^2/kg^2). calculate the orbital speed for a satellite at an altitude of 1.62x10^6 m.

v = 2917.35 m/s

Explanation:

let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.

at any point in the orbit  the forces acting on the satelite are balanced such that:

Fc = Fg

mv^2/r = GmM/r^2

v^2 = GM/r

   v = \sqrt{GM/r}  

      = \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}

      = 2917.35 m/s

Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 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

the entire statement is true and correct.  

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