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andriy [413]
3 years ago
14

Write down examples of some natural acids and natural bases. Also write use of them.

Chemistry
1 answer:
trasher [3.6K]3 years ago
3 0

Answer:

Formic acid, citric acid, Oxalic acid, washing soda, baking soda, etc. can be some examples of natural acids and natural bases. They both have domestic, industrial, and various other purposes.

Explanation:

<h3><u>NATURAL ACIDS</u>:</h3>

There are lots of natural acids present in our nature. Some of them are the following:

> <u>Formic acid</u>

 USE: It is used in the stimulation of oil and gas wells as it is less reactive towards the metal.

> <u>Citric acid</u>

 USE: It is considered as the best rust remover as it doesn't harm the metal just remove the rust.

> <u>Oxalic acid</u>

USE: It easily remove iron and ink stains and that's why it is used as an acid rinsing material in Laundries.

<h3><u>NATURAL BASES</u>:</h3>

There is a variety of natural base found in our nature which founds a lot of uses in day to day life. some of them are the following:

> <u>Washing soda</u>

USE: It is used in commercial detergent mixture to treat hard water.

> <u>Baking soda</u>

USE: It is the best rising agent used mostly in cooking and for domestic purposes like removing stains, etc..

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The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer
sergey [27]

Answer:

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Explanation:

The pH of a buffer solution is calculated using following relation

pH=pKa+log(\frac{salt}{acid} )

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.

pKa = -log [Ka]

For HC₃H₅O₃

pKa = 3.1

For CH₃NH₃⁺

pKa = 10.64

pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)

A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

8 0
3 years ago
How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC
Reptile [31]

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

4 0
3 years ago
When a bond is formed, energy is _ the atoms
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"released from" the atoms
5 0
3 years ago
What mass of lead (density 11.4 g/cm3) would have an identical volume to 20.1 g of mercury (density 13.6 g/cm3)?
oksano4ka [1.4K]
In order to determine the mass of lead, Pb, that would have an identical volume as 20.1 g of mercury, Hg, we first get the equivalent volume of 20.1 g of mercury. This can be done by using the given density of mercury which is equal to 13.6 g/cm^3. This is shown in the following equation:

20.1 g Hg/ 13.6g/cm^3 = 1.4779 cm^3 of Hg

Consequently, we use this obtained volume which is the target volume for lead. Again, we make use of the density of the substance, in this case, lead has a density of 11.4 g/cm^3. The resulting equation is then:

11.4 g/cm^3 x 1.4779 cm^3 = 16.8485 g of Pb

Thus, the amount of lead that has an identical volume as 20.1 g of mercury is 16.8485 g Pb.   
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