A compound contains 28.6 magnesium 14.3 carbon 57.1 oxygen what is the empirical formula

Option 3

Explanation:

These are the steps for empirical forumla

1) If given percentage, assume everything in grams

2) Convert those grams to moles (using specific molar mass of each unique element)

3) Divide all the moles by the smallest mole you discovered.

4) Answer will be close to a whole number for each element and that's the answer.

First step, 28.6 g Mg, 14.3 g C, 57.1 g O

Second step:

28.6/24.305 = 1.1767 Mg

14.3/12.01= 1.1907 C

57.1/16.00 = 3.5688 O

Third step:

1.1767/1.1767 = 1 Mg

1.1907//1.1767 ~ 1 C

3.5688/1.1767 ~ 3 O

Put everything together you get!

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