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Airida [17]
3 years ago
11

How do we separate sugar solution with sand

Chemistry
2 answers:
mihalych1998 [28]3 years ago
6 0
You can separate both sand and sugar by adding water because sugar will dissolve sand cannot
Elina [12.6K]3 years ago
3 0
A sugar solution with sand can be separated by dissolving the solution into water, then the sugar will be dissolved with the water and the sand will stay on the bottom of the water.
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Why does CCl4 dissolve in iodine
nirvana33 [79]

Answer: They are both non polar molecules compounds although water is a polar molecule compound. There are no significant attractions.

8 0
3 years ago
Can someone please explain osmosis????​
ANEK [815]

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷Osmosis is a process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concentrated solution into a more concentrated one, thus equalizing the concentrations on each side of the membrane.

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3 0
4 years ago
Read 2 more answers
A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de
jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0
4 years ago
A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

4 0
4 years ago
how much potassium bromide, in grams, should be added to water to prepare 0.50 of solution with molarity of 0.125M?
Flura [38]
Volume = 0.50 L

Molar mass  KBr = 119.002 g/mol

Molarity = 0.125 M

Mass ( KBr) = ?

 mass = molarity * molar mass * volume

mass = 0.125 * 119.002 * 0.50

mass = 7.437625 g of KBr

hope this helps!
5 0
3 years ago
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