ompute the approximate acceleration of gravity for an object above the earth's surface, assigning accel_gravity with the result. The expression for the acceleration of gravity is: (G * M) / (d 2), where G is the gravitational constant 6.673 x 10-11, M is the mass of the earth 5.98 x 1024 (in kg), and d is the distance in meters from the Earth's center (stored in variable dist_center). Sample output for the given program is:

Explanation:

Given that

Gravitational Constant, G = 6.673*10^-11

Mass of the earth, M = 5.98*10^24 kg

Since we are no given a distance, d. It's safe for us to assume any distance we like.

For example, if we assume our d to be 1 m, then we have

a = GM/d²

a = (6.673*10^-11 * 5.98*10^24) / 1²

a = 3.99*10^14 m/s²

In another instance, if we assume the distance d, to be the distance between the Earth and the Sun, we have that d to be 150*10^9 m. Our acceleration would then be

a = GM/d²

a = (6.673*10^-11 * 5.98*10^24) / (150*10^9)²

a = 3.99*10^14 / 2.25*10^24

a = 1.77*10^-8 m/s²

We have two different accelerations

Thus we can see that our acceleration would be dependent on the distance given

  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

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: