"a 10.0 ml sample of this solution was transferred to a 500.0-ml volumetric flask and diluted to the mark with water. then 10.0 ml of the diluted solution was transferred to a 250.0-ml flask and diluted to the mark with water. what is the final concentration of the kno3 solution? "

If the original solution is "X" M in KNO₃ (moles/Lt), then final solution will be in concentration (2/250)×X M

Explanation:

Original solution of KNO₃ is X M, then, we take a sample of 10 ml and transferre it to 500 ml:

1000ml=X moles KNO₃

10ml=1/100 X moles KNO₃, that will be in 500ml

We then take 10 ml of this new solution and transferre it to a flask with a final volume of 250ml, so:

500ml=(1/100) X moles KNO₃

10ml=(1/5000) X moles KNO₃, these moles will now be in 250 ml

With this, we can calculate the final concentration (in M, or moles/Lt):

250ml=(X/5000) moles KNO₃

1000ml=2/250×X M KNO₃

So, once you know the original concentration of KNO₃ solution, you should just multiply this number by (2/250) to get final concentration of 250 ml solution obtained by multiple dilutions

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

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