Hydrocarbons are compounds that contain

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0

3 years ago

How much heat is required to vaporize 25g of water at 100*c

ElenaW [278]

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

4 0

3 years ago

A 3.82 g piece of limestone contains 2.62 g of CaCO3

Kobotan [32]

Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.

<h3>What is mass percentage</h3>

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$mass percentage=\frac{mass of solute}{mass of solution}x100$