Two sample of carbon tetrachloride are decomposed into their constituent elements. one sample produces 32.4 g of carbon and 373 g of chlorine, and the other sample produces 12.3 g of carbon and 112 g of chlorine. are these results consistent with the law of constant composition?

These results are not consistent with the law of constant composition.

Explanation:

Law of constant composition states that a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

Following this definition, for carbon tetrachloride (CCl₄), the ratio of its component elements by mass is:

Mass of Cl / Mass of C

In the first sample, mass of Cl is 373g and mass of C is 32,4. The ratio is:

373g/32,4g = 11,5

In the second sample, mass of Cl is 112g and mass of C is 12,3. The ratio is:

112g/12,3 = 9,11

As the ratio of its component elements is not fixed, it is possible to say that these results are not consistent with the law of constant composition.

I hope it helps!

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

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.