All categories

2 years ago

How does an indicator differentiate between an acid and base?

Chemistry

1 answer:

Svet_ta [14]2 years ago

8 0

Answer:

that is all I know sorry hope that may help you :)

Explanation:

Indicator is a substance which shows different colors in acidic and basic medium. Indicators can be natural (derived from natural sources) or artificial ( man-made) for example :

*Litmus is an indicator. Acid turns blue litmus into red while base turns red litmus into blue

* Turmeric solution does not show change in color (remains yellow) in acidic solution and turns red in basic solution.

You might be interested in

You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted

klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water = 600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = $\frac{C_1}{C_2}$

where; $C_1=$ solubility of compound in the organic solvent and $C_2$ = solubility in aqueous water.

So; we can represent our data as:

$K=(\frac{A_{(g)}}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL} )$

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6= $(\frac{0.6-B(mg)}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL})$

4.6 = $\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}$

4.6 × $(\frac{B_{(mg)}}{100mL})$ = $(\frac{0.6-B(mg)}{60mL} )$

4.6 B $*\frac{60}{100}$ = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = $\frac{0.6}{3.76}$

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0

3 years ago

Read 2 more answers

A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0

3 years ago

PLEASEEEEEEEEEEEE HELPPPPPPPP I BEGGGGG FOR HELPPPPP

Elza [17]

Answer: There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles$

1 mole of $H_2O$ contains = $6.023\times 10^{23}$ molecules

Thus 3.5 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23}$ molecules.

There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

3 0

3 years ago

The chart shows the solubility of different substances.

Annette [7]

Answer: 1) Temperature can change the solubility of a solute.

Explanation:

The chart is missing so there is no way to tell what does the graph show.

Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..

1) Temperature can change the solubility of a solute.

Yes, temperature definetly can, and mostly do, modify the solubility of a solute.

You can search any chart of solubility and will find that.

I can give you two examples:

a) Sodium chloride: dissolve some spoons of salt in a cold water until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.

b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.

The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).

2) Temperature has no affect on the solubility of a solute.

Since this is the opposite to the first statement and the first is true, this is false.

3) Salt has a greater solubility than sugar.

False.

This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.

That is something you ca see in a chart or you can prove by yourself.

4) Nitrite salt has a greater solubility than sugar. 

False.

Looking at some data you can find that sodium nitrite solutiliby is aroun 70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.

8 0

3 years ago

Read 2 more answers

Refrigerators are usually kept at about 5°C, while room temperature is about 20°C. if you were to take an empty sealed 2 liter s

MakcuM [25]

Answer:

c) No, because Celsius is not an absolute temperature scale

Explanation:

converting 5 oC to kelvin which is the absolute temperature scale gives = 273 + 5 = 278 K

and converting 20 oC to kelvin = 20 + 273 = 293 K

the ratio = 278 / 293 = 0.94 approx 1 not 4

8 0

3 years ago