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1 year ago

8

39. suppose an aqueous solution contains both table sugar and table salt. can you separate either of these solutes from the wate

r by filtration? explain your reasoning.

1 answer:

Degger [83]1 year ago

7 0

Answer:

No, assuming that the salt/sugar is already dissolved

Explanation:

As long as the particle size is too big, it won't filter through. Therefore, if it is dissolved, it will pass through the filter.

If you were to throw rocks in there or something, and they are non-dissolvable, then yes.

You might be interested in

Determine the number of moles in 382.5g of CO-2

exis [7]

Answer: 8.691 mols of CO₂

Explanation:

To find the number of moles in a given grams, you want to use the molar mass.

Let's first find the molar mass of CO₂.

Carbon's molar mass is 12.011 g/mol

Oxygen's molar mass is 15.999 g/mol

To find molar mass of CO₂, we want to add up the molar mass of carbon and oxygen. Remember, there are 2 Oxygens so we need to mulitply that by 2.

12.011+2(15.999)=44.009 g/mol

Now that we have molar mass, we can convert 382.5 g to mols.

$382.5g*\frac{1 mol}{44.009g} =8.691 mol$

There are about 8.691 mols of CO₂.

5 0

3 years ago

Drupady [299]

What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that

does its temperature vary by 25 ° C?

Answer:

143.75cal

Explanation:

Given parameters:

Mass of steel = 50g

Specific heat capacity of the steel = 0.115cal/g°C

Temperature = 25°C

Unknown:

Amount of heat = ?

Solution:

The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.

Amount of heat = m C (ΔT)

m is the mass

c is the specific heat capacity

ΔT is the temperature change

Now insert the parameters and solve;

Amount of heat = 50 x 0.115 x 25

Amount of heat = 143.75cal

4 0

2 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

2 years ago

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr

lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Using the equation,

Δ $T_{f}$ = i $K_{f}$ m

where:

Δ $T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ $T_{f}$ = i $K_{f}$ m we get,

Δ $T_{f}$ = 2 × 1.86 × 0.1219

Δ $T_{f}$ = 0.4536°C

So, Δ $T_{f}$ of solution is,

Δ $T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ $T_{f_{solution} }$ = - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Learn more about freezing point here;

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7 0

1 year ago

I was born in ……… west.<br><br>complete this sentense with article

Xelga [282]

Answer:

Article is

I was born in <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em> west .

4 0

2 years ago