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IgorC [24]
3 years ago
14

Select the fastest way to differentiate between the organic and aqueous layers of an extraction.a. Hold a white piece of paper b

ehind the separatory funnel. The organic layer will have a slight yellow color, while the aqueous layer will be colorless.b. Add a few drops of one of the layers to a test tube containing 1 mL of water. Shake the test tube to determine the solubility of the layer in water.c. Separate the layers and heat them. The organic layer will have the lowest boiling point.d. Weigh equal volumes of each layer. The aqueous layer will have the highest mass.
Chemistry
1 answer:
valentinak56 [21]3 years ago
6 0

Answer:

b. Add a few drops of one of the layers to a test tube containing 1 mL of water. Shake the test tube to determine the solubility of the layer in water

Explanation:

Option a is not true, it depends on the compound being extracted.

Option c is not true, although most of the solvents used in extractions have lower boiling point than water there are exceptions, for example toluene.

Option  d is not true. Again most of the solvents used in extractions are less dense than water, there are many exceptions, for example chloroform, so for equal volumes the chloroform layer will weigh more.

Option b. is the correct one.

One will test the  miscibility of the layer in water. If it inmiscible then one would know is the organic layer. If it is the aqueous layer then it will completely be miscible.

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A sample of gas occupies 10 L at STP. What
puteri [66]

Pressure is 5.7 atm

<u>Explanation:</u>

P1 = Standard pressure = 1 atm

P2 = ?  

V1 = Volume = 10L

V2= 2.4L

T1 = 0°C + 273 K = 273 K

T2 = 100°C + 273 K = 373 K

We have to find the pressure of the gas, by using the gas formula as,

$\frac{P 1 V 1}{T 1}=\frac{P 2 V 2}{T 2}

P2 can be found by rewriting the above expression as,

$P 2=\frac{P 1 \times V 1 \times T 2}{T 1 \times V 2}

Plugin the above values as,

$P 2=\frac{1 \text {atm} \times 10 L \times 373 \mathrm{K}}{273 \mathrm{K} \times 2.4 \mathrm{L}}=5.7 \text { atm }

4 0
3 years ago
1. A neutral atom of fluorine (F) has 9 electrons and an electron configuration of 1s2 2s2 2p5. How will it ionize to achieve an
Natasha2012 [34]
These answers dont make sense

1. the 2s2 orbital will give one of its electrons to the 2p5 orbital so the configuration would be 1s22s12p6 (2s1 is half filled and 2p6 is completely filled which is a much more stable configuration)

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3 years ago
Identify the principle or law that identifies the relative age of the intrusion compared to layer Q.
Darya [45]

Answer:

The answer is C (law of superposition)

Explanation:

7 0
3 years ago
A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i
almond37 [142]

The half reaction that occurs at the Au electrode is  1.64

<u>Explanation:</u>

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e-  ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e-  ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF  log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1  0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is  1.64

<u />

6 0
3 years ago
Hybrid orbitals of the sp3d type occur in sets of four. True or False?
lara31 [8.8K]

well , it's true because they are of sp3d type occur on sets of four

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3 years ago
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