When a 7.40-mg sample of a compound containing carbon is burned completely, 18.6 mg of carbon dioxide is produced. What is the mass percentage of carbon in the compound?

68.6 % of the compound is carbon

Explanation:

Step 1: Data given

Mass of 7.40 mg = 0.0074 grams

Mass of carbon dioxide = 18.6 mg = 0.0186 grams

Molar mass CO2 = 44.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.0186 grams / 44.01 g/mol

Moles CO2 = 4.23*10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 4.23*10^-4 moles CO2 we have 4.23*10^-4 moles C

Step 4: Calculate mass C

Mass C = 4.23*10^-4 moles * 12.01 g/mol

Mass C = 0.00508 grams

Step 5: calculate the mass percentage of carbon

% C= (0.00508 grams / 0.0074 grams ) * 100 %

% C = 68.6 %

68.6 % of the compound is carbon

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

the answer would be b i believe