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Nata [24]
3 years ago
14

A student has a flask containing two immiscible liquids. One of the liquids is a solution of a solid in water. Describe how you

would separate the mixture into three separate components.
Chemistry
2 answers:
Oksanka [162]3 years ago
6 0

The mixture can be separated into three separate components by using a centrifuge and a separating funnel.

A centrifuge is used to separate a solid that is held in suspension by a liquid. In this case a solid dissolved in water. In contrast, a separating funnel is used to separate two immiscible liquids.

-First take the flask containing the two immiscible liquids and pour contents in to a closed separating funnel.

- The most dense liquid will settle at the bottom, then open the separating funnel valve to collect contents of dense liquid into a flask.

- The remaining liquid on the separating funnel can be collected in a different flask.

-Now take the liquid that is a solution of a solid in water and pour in a test tube. Place the contents in a centrifuge.

- Centrifuge at high speed until the solid sinks to the bottom of the test tube, then decant the liquid into a flask and the solid remains in the test tube

I am Lyosha [343]3 years ago
3 0

Answer:

It can be separated using a separating funnel followed by crystallization.

Explanation:

First the mixture is put into a separating funnel. This will separate the two immiscible liquids. When the two immiscible liquids are separated, we recall that one is a solution of a solid solute and we wish to recover the solid from the solution whilst recovering the solvent also.

The solid can only be recovered by crystalization if we do not want to loose the solvent. Seeds of already crystalized solute may be added to system to induce crystalization.

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Gold is alloyed with other metals to increase its hardness in making jewelery.a) Consider a piece of gold jewelry the weighs 9.8
irinina [24]

Answer:

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

Explanation:

Mass of jewellery = 9.85g Volume of jewellery = 0.675cm^3.

Density of gold = 19.3g/cm^3 and density of silver = 10.5g/cm^3

Let the volume of gold in the jewellery be X and the volume of silver in the jewellery be Y

Hence we have

Density = mass/volume or mass = volume × density = for gold = X × 19.3g/cm^3 and for the silver = Y × 10.5g/cm^3

19.3X +10.5Y = 9.85g

Also volume of jewellery is given by

Volume of silver in the jewellery + volume of gold in the jewellery = 0.675cm^3.

X + Y = 0.675cm^3.

Solving the above equations we have

Y = 0.675 - X

Which gives

19.3X + 10.5Y = 9.85g

19.3X + 10.5 × (0.675 - X) =9.85g

19.3X + 7.0875 - 10.5X = 9.85

8.8X + 7.0875 = 9.85

8.8X = 2.7625

or X = 0.3139 cm^3

But Y = 0.675 - X

Hence Y = 0.675 - 0.3139 = 0.3611 cm^3

Mass of gold in the jewellery = volume of gold × Density of gold = 0.3139 × 19.3 = 6.059 g

Also mass of silver = 10.5 × 0.3611 = 3.7913g

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

7 0
3 years ago
I hate chemistry but someone help please, no links or you’ll be reported
Tpy6a [65]

Answer:

A, D, E

Explanation:

4 0
3 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
If 16.0 mL of acetone is dissolved in water to make 155 mL of solution what is the concentration expressed in volume/volume % of
Lady_Fox [76]
Percentage by volume of solution is the percentage volume of solute in total volume of solution.
Volume percentage (v/v%) = volume of solute / total volume of solution x 100%
volume of solute - 16.0 mL
total volume of solution - 155 mL 
v/v% = 16.0 / 155 x 100% = 10.32%
this means that in a volume of 100 mL solution, 10.32 mL is acetone.
7 0
3 years ago
In the following overall chemical reaction, aluminum displaces chromium from chromium(III) oxide and forms aluminum oxide.
mafiozo [28]

Answer:

Change in enthalpy for the reaction is -536 kJ

Explanation:

  • Overall chemical reaction can be represented a summation of two given elementary steps with slight modification.
  • Take reaction (1a) and divide stoichiometric coefficients by 2
  • Take reverse reaction (2a) and divide stoichiometric coefficient by 2
  • Then add these two modified elementary steps to get overall chemical reaction
  • \Delta H is an additive property. hence value of \Delta H will be changed in accordance with modification

2Al+\frac{3}{2}O_{2}\rightarrow Al_{2}O_{3}.....\Delta H_{1}=\frac{-3351.4}{2}kJ=-1675.7kJ

Cr_{2}O_{3}\rightarrow 2Cr+\frac{3}{2}O_{2}....\Delta H_{2}=\frac{2279.4}{2}kJ=1139.7kJ

--------------------------------------------------------------------------------------------------------------

2Al+Cr_{2}O_{3}\rightarrow 2Cr+Al_{2}O_{3}....\Delta H=\Delta H_{1}+\Delta H_{2}=-1675.7+1139.7kJ=-536kJ

6 0
3 years ago
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