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vova2212 [387]
3 years ago
13

You have a cup of hot water and you dissolve as much salt in it as possible. You then put the cup into the refrigerator and you

check it again after 24 hours. What would you notice when you checked the cup after 24 hours in the refrigerator? None of the salt would settle out of the water. Some of the salt would settle out of the water. All of the salt would settle out of the water. The salt water would have changed colors.
Chemistry
2 answers:
n200080 [17]3 years ago
7 0

Answer:

Some of the salt would settle out of the water.

Explanation:

This is an example of a supersaturated solution.

In this example, the salt is the analyte, and the water is the solvent.

When you mix the salt into the water in an ambient temperature there is a point where you can't solve more salt, because the amount of water isn't able to solubilize any more quantity of analyte. This is named supersaturation point.

If you expose the mix to the heat, the energy supply to the solution make that the excess of salt can be solubilized, but this effect only last until the heat still going on.

So when you put the solution into the refrigerator, the quick down of the heat makes that the excess of salt settles out of the water.

And all this process occurs because the solution always wants to reach the equilibrium between the analyte and the solvent.

hoa [83]3 years ago
6 0
Some of the salt would settle out. When the water was heated, it was able to absorb more salt than usual. This is known as super saturation. When the water is frozen it cannot hold as much salt, so some of it has to come out.
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Which of the following is an irreversible physical change?
Yuliya22 [10]

Answer:

b) sharpening a pencil

Explanation:

If you melt lead, boil water, or dissolve sugar in water, you can return all of them back to their original state. If you sharpen a pencil, you can't reattach the shavings as they were originally.

3 0
4 years ago
Why do you require an acid catalyst to make an ester? Why not just mix acid and alcohol? Describe an alternate method of making
djverab [1.8K]

Answer:Acid catalyst is needed to increase the electrophilicity of Carbonyl group of Carboxylic acid as alcohol is a weak nucleophile.

Alternatively esters can be synthesised by converting carboxylic acid into acyl chloride using thionyl chloride(SOCl_{2} and then further treating acyl chloride with alcohol.

Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

Alcohol and esters can also be distinguished using IR as alcohols would contain a broad intense peak at around 3200-3600cm_{-1}

Explanation: For the synthesis of esters using alcohol and carboxylic acid we need to add a little amount of acid in the reaction . The acid used here increases the electrophilicity of carbonyl carbon and hence makes it easier for a weaker nucleophile like alcohol to attack the carbonyl carbon of acid.

The oxygen of the carbonyl group is protonated using the acidic proton which  leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

If we simply mix the acid and alcohol then no appreciable reaction would take place between them and ester formation would not take place because the carboxylic acid in that case is not a good electrophile whereas alcohol is also not a very strong nucleophile which can attack the carbonyl group.

Alternatively we can use thionyl chloride or any other reagent which can convert the carboxylic acid into acyl chloride. Acyl chloride is very elctrophilic and alcohol can very easily attack the acyl chloride and esters could be synthesized.

The carboxylic acid and ester can very easily be distinguished on the basis of broad intense OH stretching frequency peak at around 2500-3200cm_{-1} . The broad intense OH stretching frequency peak is present in carboxylic acids as they contain OH groups and absent in case of esters .

Likewise esters and alcohols can also be distinguished on the basis IR spectra as alcohols will have broad intense spectra  at around 3200-3600cm_{-1}corresponding to OH stretching frequency whereas esters will not have any such peak. Rather esters would be having a Carbonyl stretching frequency at around 1720-1760

4 0
3 years ago
Given the setup of the equilibrium constant expression, a reaction that has a high equilibrium constant will "favor" the product
ss7ja [257]

Answer:

for points

Explanation:

loll

5 0
3 years ago
Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

7 0
3 years ago
Given 20.0 g of (NH4)3 PO3 how many moles is this? How many formula units is it? How many hydrogen atoms are present?
Setler79 [48]

Answer:

no of moles = (mass) ÷ molar mass

you get the molar mass from the periodic table

n = 20.0g ÷ ( 149.09 g/mol )

= 0.134 mol

number of formula units = number of moles × (6.022×10^23)

= 8.07×10^22 units

number of hydrogen atoms in (NH4)3 PO3 = 12 × number of formula units

= 9.68×10^23 hydrogen atoms

8 0
3 years ago
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