The water in a 150-mL flask is poured into a beaker that contains 600 mL of ethyl alcohol. Which substance is the solvent in this mixture?

Answers

Answer 1
Answer:

Answer:

Ethyl alcohol is a solvent on the mixture.

Explanation:

Solution: It is a mixture of liquid in which solute (minor component) is distributed uniformly in a solvent (major component).

In a solution, mass of solute is always less than the mass of solvent.

Mass of water = 150 mL

Mass of ethyl alcohol = 600 mL

Mass of water < Mass of ethyl alcohol

Ethyl alcohol is a solvent on the mixture.

Answer 2
Answer: The solvent in the mixture is the ethyl alcohol assuming that the ethyl alcohol is not very dilute. The amount of the solvent must be greater than the amount of the solute. If the ethyl alcohol is dilute such that the resulting solution will have more water than pure ethyl alcohol, the solvent is water and the solute is ethyl alcohol.

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The equation below shows the decomposition of lead nitrate. How many molecules of PbO are produced when 11.5 g NO2 is formed?2Pb(NO3 )2 (s) → 2PbO(s) + 4NO2 (g) + O2 (g)

Answers

Answer is in the file below

An estser is the reaction product between: a. Alcohol and organic acid B. Alcohol and oxygen
C. Ketones and Oxygen.
D. Aldehydes and Alcohol

Answers

An ester is the reaction product between A. alcohol and organic acid. 

Convert 22.0 mL to L

Answers

0.022


One milliliter is equal to a thousandth of a liter, so divide by 1000..

Final answer:

To convert 22.0 mL to L, divide the given value by 1000.

Explanation:

To convert 22.0 mL to L, you need to use the conversion factor of 1 L = 1000 mL.

Therefore, to convert mL to L, divide the given value by 1000.

So, 22.0 mL = 22.0 mL / 1000 = 0.022 L.

Learn more about Unit Conversion here:

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46 What occurs at one of the electrodes in both an electrolytic cell and a voltaic ccl1?( I) Oxidation occurs as electrons arc gained at the cathode.
(2) Oxidation occurs as electrons are lost at the anode.(3) Reduction occurs as electrons are gained at the anode.
(4) Reduction occurs as electrons are lost at the cathode.

Answers

Oxidation occurs as electrons are lost at the anode at one of the electrodes in both an electrolytic cell and a voltaic cell. The answer is number 2. The rest of the choices does not answer the question given above.

A student needs .575 moles of sodium chloride for an experiment. how many grams she she mass out

Answers

The molecular weight of NaCl is 58.44 g/mol.
0.575 mol * 58.44 g/mol = 33.6 grams of NaCl

An evaporation-crystallization process is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fresh feed to the process contains 19.6 wt% K₂SO₄. The wet filter cake consists of solid K₂SO₄ crystals and a 40.0 wt% K₂SO₄ solution, in a ratio 10.0 kg crystal/kg solution. The filtrate, also a 40.0 wt% K₂SO₄ solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0% by weight is evaporated. The evaporator has a maximum capacity of 175. kg water evaporated/s. a) Assume the process is operating at maximum capacity. Draw and label a flowchart and do the degreeof-freedom analysis for the overall system, the recycle-fresh feed mixing point, the evaporator, and the crystallizer.
b) Calculate the maximum production rate of solid K₂SO₄, the rate at which fresh feed must be supplied to achieve this production rate, and the recycle ratio kg/h recycled stream to kg/h fresh feed stream.

Answers

Answer:

From the degree of freedom analysis, the degree of freedom of the system and its components are equal to zero hence the system is well defined

Explanation:

 

To appraise the evaporation-crystallisation process, we go over the system to check if it is well defined from the available information as follows

45% by weight of inlet water is evaporated hence where inlet consists of 19.6% by weight of K₂SO₄ we have, Molar mass of K₂SO₄ = 174.259g/mol. Thus for every mole of K₂SO₄, we have 174.259×100/19.6 = 889.1g  of solution is fed per mole of K₂SO₄,

Also the stream of concentrate leaving the evaporator contains

889.1 – 174.259 = 714.7 grams of water, and if 45% by weight of water is evaporated we have

45% of 889.1 is evaporated leaving a solution of weight = 889.1 × 55/100 = 489grams of solution which contains

100×174.259÷489 or 35.6% by weight of K₂SO₄ concentrate leaving the evaporator and moving on to the evaporator

However, 175. kg of water is evaporated/s hence from the previous calculation, quantity of water per mole of K₂SO₄ evaporated = 0.45×889.1= 400.1g which in comparison with actual quantity gives mass flow into

0.4001/175 = 0.003 or 437.39 to 1 hence the mass flow rate is 889.1g×437.39 = 388884g/s or 388.9Kg/s

 

 

a. Degrees of freedom analysis for the overall system

We have the following 4 unknowns in the overall system viz

m1, m3, m4 and m5

where m1 = maximum mass flow rate into the system

m3 = mass rate of evaporated water

m4 = maximum mass of solid K₂SO₄ crystals produced from the crystallizers

m5 = recycle ratio

While we have

1) Information, maximum capacity of evaporation from where we can calculate the maximum rate of feed supply

2) information, including chemical formula, to determine the maximum production rate

3) Information to calculate the water evaporated from fraction of water which is evaporated to that which is supplied

4) information to calculate the recycle ratio

Hence degrees of freedom = 4 – 4 =0

 

b. Degree of freedom analysis for the recycle-fresh feed mixing point

For the recycle-fresh feed mixing point we have m1 and m5, two unknowns

Where m1 is maximum feed rate and m5 is the mass of filtrate and we are given the compound molecular formula and the maximum flow rate from where we can calculate both m1 and m5

Hence the degrees of freedom = 0

 

c. Degree of freedom analysis for the evaporator

 

For the evaporator we have three unknowns m1,m2 and m3 and the available information are

1. The maximum water processing capacity of the evaporators and

2.   The percentage quantity of water evaporated

Which is 2 hence we have 2 – 2 = 0 degrees of freedom

 

and

d. Degree of freedom analysis for the crystallizer the unknowns are m2, m4, m5

 

For the crystallizer the unknowns are m2, m4, m5The information available are

1. the ratio of crystals per kilogram of solution

2. The concentration of the recycled K₂SO₄ solution

3. Information of the maximum capacity of the evaporator so as to calculate the mass of concentrates leaving the evaporator and moving towards the crystallizer

Hence, we have 3 -3 = 0 degrees of freedom

 

From the degree of freedom analysis, the degree of freedom of the system and its components are zero hence the system is well defined