A 72 kg athlete climbs a rope to a height of 12m. Calculate the increase in gravitational potential energy it has experienced.
Answer:
8467.2J
Explanation:
Given parameters:
Mass of the athlete = 72kg
Height of the climb = 12m
Unknown:
Increase in gravitational potential energy it has experienced = ?
Solution:
Gravitational potential energy is the energy due to the position of a body. It is mathematically expressed as;
Gravitational potential energy = m x g x h
m is the mass
g is the acceleration due to gravity = 9.8m/s²
h is the height
Insert the parameters and solve;
Gravitational potential energy = 72 x 9.8 x 12
GPE = 8467.2J
Noble gases:) they are very non-reactive
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:

Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n



According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:

Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)