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3 years ago

Which quantities can be measured

Chemistry

2 answers:

chubhunter [2.5K]3 years ago

4 0

Explanation:

Any quantity which can be measured is called a physical quantity.

babymother [125]3 years ago

3 0

Explanation:

Physical quantity can be measured.

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A high jumper jumps 2.09m. If the jumper has a mass of 72kg, what is his gravitational potential energy at the highest point of

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3 years ago

A flame is the result of a chemical reaction that releases energy. Do you think a similar kind of chemical reaction is happening

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2 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

Katen [24]

Answer:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

Explanation:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

$Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}$

The pKa is

$-Log (Ka) = -Log (6.2x10^{-8}) = 7.2$

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.

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3 years ago

A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6

joja [24]

<u>Answer:</u> The concentration of $CH_3COOH$ comes out to be 0.16 M.

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $CH_3COOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=25mL\\n_2=1\\M_2=0.175M\\V_2=26.6mL$

Putting values in above equation, we get:

$1\times M_1\times 25=1\times 0.175\times 26.6\\\\M_1=0.1862M$

Hence, the concentration of $CH_3COOH$ comes out to be 0.1862 M.

4 0

3 years ago

Which quantities can be measured​

Which quantities can be measured