6th grade help me plzzzz

What happened to the mixture as it is poured into the strainer with cloth?

zzz [600]

Answer:

The solid particles will be trapped behind on the strainer while the liquid molecules will pass or filter through.

Explanation:

Filtration is a process of separation of mixtures involving an insoluble solid and a fluid such as liquids or gases using a filtering medium such as paper or cloth or any other material. During the process, the liquid molecules pass through the filtering medium whereas the solid particles are not allowed to pass through but are retained by the medium. The reason is that the pores on the filtering medium are only small enough for the liquid molecules to pass through but not the solid particles.

An example of a mixture that can be separated through filtration is a suspension of chalk and water. Also, a mixture of water and solid particles such as water and tea leaves can also be separated using a strainer with cloth.

In the question, if a mixture of insoluble solids in a liquid is passed through a strainer with cloth, the solid particles will be trapped behind on the strainer while the liquid molecules will pass through.

3 0

2 years ago

For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol

Lapatulllka [165]

Answer:

$C_{vm}$ of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

$C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}$

Where:

$C_{pm}$ = Specific heat of gas at constant pressure = 75.3 J/mol·K

$C_{vm}$ = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - $C_v$ = $\frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81 \times 10^{-5} }{4.52 \times 10^{-5} }$

$C_{vm}$ = $\frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81 \times 10^{-5} }{4.52 \times 10^{-5} }$ - 75.3 = 256.834 J/mol·K.

8 0

3 years ago

A gas has a pressure of 8.5atm and occupies 24L at 25∘C. What volume (in liters) will the gas occupy if the pressure is increase

Vesna [10]

The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

Final pressure (P₂) = 13.5 atm

Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298} = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\$