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1 year ago

A mixture of sand and table salt cam be separated by filtration because the substances in the mixture differ in

1 answer:

6 0

A mixture of sand and table salt can be separated by filtration because the substances in the mixture differ in solubility. Because sand and table salt have a different ability to dissolve they can easily be separated. What can be done to separate this mixture is to add water to it and stir it thoroughly. The salt will easily dissolve in water while the sand will stay insoluble. The filtration of this mixture will result in salt being drained off with water while only the sand will remain.

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The temperature of the carbon dioxide atmosphere near the surface of venus is 475°c. calculate the average kinetic energy per mo

harkovskaia [24]

Answer: <span>9330 j/mol
</span>
The temperature of the gas is 475 ° Celcius which is equal to: 475 +273= 748 °K. The formula for kinetic energy of individual atoms would be

K= 3/2 * kB * T
If kB is 1.38 * 10^-23 J/K and 1 mol is made from 6.02*10^23 molecule, then the kinetic energy of 1 mol CO2 would be:
K= 3/2 * kB * T
K= 3/2 * 1.38 * 10^-23 * 748 * 6.02 *10^23 =9324 J/mol

8 0

1 year ago

Determine the transition length at the entrance to a 10 mm tube through which 100 percent glycerol at 60 C is flowing at a veloc

Readme [11.4K]

Answer: Option (d) is the correct answer.

Explanation:

The given data is as follows.

Tube diameter d = 10 mm = 0.01 m

Velocity of glycerol, v = 0.5 m/s

Density of glycerol ( $\rho$ ) = 1240 kg/m3

Dynamic viscosity of glycerol ( $\mu$ ) = 0.0813 pa.s

Reynolds number (Re) = $\rho \times velocity \times \frac{density}{\mu}$

= $1240 \times 0.5 \times \frac{0.01}{0.0813}$

= 76.26

Therefore, according to Reynolds number we can say that flow is laminar.

Lt = $0.05 \times Re \times d$

= $0.05 \times 76.26 \times 0.01$

= $0.03813 m$

As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to $0.03813 m \times \frac{1000 mm}{1 m}$

= 38.13 mm

Thus, we can conclude that the transition length of glycerol is 38.13 mm.

5 0

2 years ago

What is the percent yield of sodium hydroxide in the given reaction? The reaction was performed using 45 g NaHCO3 and 18 g NaOH

Vlada [557]

Answer:

The percent yield = 83.33 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $NaHCO_3$ :-

Mass of water = 45 g

Molar mass of $NaHCO_3$ = 84.007 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{45\ g}{84.007\ g/mol}$

$Moles\ of\ NaHCO_3= 0.54\ mol$

For $NaOH$ :-

Given mass = 18 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{18\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.45\ mol$

According to the given reaction:

$NaHCO_3\rightarrow NaOH+CO_2$

1 mole of $NaHCO_3$ on reaction forms 1 mole of $NaOH$

Thus,

0.54 mole of $NaHCO_3$ on reaction forms 0.54 mole of $NaOH$

Moles of $NaOH$ formed = 0.54 moles

Moles of $NaOH$ actually formed = 0.45 moles

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 0.54 moles

Experimental yield = 0.45 moles

Applying the values in the above expression as:-

$\%\ yield =\frac{0.45}{0.54}\times 100$

$\%\ yield =83.33\ \%$

<u>The percent yield = 83.33 %</u>

7 0

1 year ago

A 10g mass of krypton occupies 15.0 l at avpressure of 210 kpa. find the volume

bagirrra123 [75]

From Boyle's law, we can use the equation
P1V1 = P2V2
to find the volume when the pressure is increased to 760 kPa:
V2 = P1V1 / P2
V2 = (210 kPa)(15.0 L) / 760 kPa
V2 = 4.14 Liters
Therefore, the volume is 3.99 Liters after the pressure increased to 760 kPa.

But if the pressure is increased to 790 kPa, the volume of the krypton will decrease to
V2 = P1V1 / P2
V2 = (210 kPa)(15.0 L) / 790 kPa
V2 = 3.99 Liters

5 0

1 year ago

What’s the mass of 0.0485 moles of Na2CO3

Artemon [7]

Answer:5.1g

Explanation:number of moles of Na2co3 is 0.0485 and it molar mass = 23×2+12+16×3=106

n=m/M

Therefore m= n×M

m= 0.0485×106

=5.141g

8 0

2 years ago