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nekit [7.7K]
2 years ago
12

Why is a quantitative observation more useful than a non-quantitative one? Which of the following are quantitative?

Chemistry
1 answer:
Strike441 [17]2 years ago
7 0

Explanation:

An observation that can be counted is known as quantitative observation. On the other hand, an observation that does not contain any numerical data is known as non-quantitative observation.

Therefore, given observations are classified as follows.

(a)  The Sun rises in the east - It is a non-quantitative observation as it cannot be measured.

(b)  A person weighs one-sixth as much on the Moon as on Earth - It is a quantitative observation as it can be measured because we know the mass of moon. Hence, we can find out the weight of person.

(c) Ice floats on water - It is a non-quantitative observation as it cannot be measured.

(d)  A hand pump cannot draw water from a well more than 34 ft deep _ It is a quantitative observation because it is measure that more than 34 ft deep water cannot be drawn by the pump.

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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
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Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

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