How many grams of manganese (IV) oxide are needed to make 5.6 Liters of a 2.1 solution? (solve for moles then convert to grams)

We are Given:

Final volume of the solution = 5.6 L

Concentration of the solution = 2.1 M

Moles of Manganese (IV) Oxide required:

Molar mass of Manganese (IV) Oxide = 87 grams/mol

since the concentration is 2.1 M, we will need 2.1 moles of Manganese Oxide for every 1L of the solution

Number of moles required = Molarity * Volume required

Number of moles required = 2.1 * 5.6 = 11.76 moles

Mass of Manganese (IV) Oxide required:

mass = molar mass * number of moles

Mass = 11.76 * 87

Mass = 1023.12 grams of Manganese (IV) Oxide

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j