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madreJ [45]
4 years ago
8

Which is the best acid to use in the preparation of a buffer with ph = 3.3?

Chemistry
1 answer:
Stella [2.4K]4 years ago
6 0

A combination of a weak acid and its conjugate base which is a mix of a weak base and its conjugate acid is so-called a buffer solution or a buffer. Buffer solutions repel a change in pH when small amount of a strong acid or a strong base are additional. Therefore, the best acid to use in the preparation of a buffer with pH=3.3 is HNO₂(Ka = 4.5 * 10 ⁻⁴). The pH of a buffer is contingent on the ration base / acid relatively than on the particular concentration of a precise solution. The precise ratio of the base to the acid for an anticipated pH can be determined from the Ka value and the Henderson HasselBalch equation.

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The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.9% H, and 31.4% O. What’s the empirical
blsea [12.9K]
To find the empirical formula you would first need to find the moles of each element:

58.8g/ 12.0g = 4.9 mol C

9.9g/ 1.0g = 9.9 mol H

31.4g/ 16.0g = 1.96 O

Then you divide by the smallest number of moles of each:

4.9/1.96 = 2.5

9.9/1.96 = 6

1.96/1.96 = 1

Since there is 2.5, you find the least number that makes each moles a whole number which is 2.

So the empirical formula is C5H12O2.
6 0
3 years ago
When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
3 years ago
Briefly discuss interpretations of your observations and results. Discuss how your observations illustrated LeChatelier's princi
Dvinal [7]

Answer:

Le Chatelier's principle can be applied in explaining the results

Explanation:

According to Le Chatelier's principle, when a constraint such as a change in concentration in this case is imposed on a chemical system in equilibrium, the system will adjust itself in such a way as to annul the constraint imposed.

Hence, when the color of the solution was more like that of the control, the reaction would shift towards the left. Similarly, when the color was more like it was towards the reactant, the reaction would shift towards the right.

If we were to prepare calcium oxalate, we should prepare it in a base solution. This is because when the base was added to calcium oxalate, it did not form any precipitate but when an acid was added to the calcium oxalate, it formed a precipitate.

4 0
3 years ago
A 10 ft diameter by 15 ft height vertical tank is receiving water (p 62.1 lb/ cu ft) at the rate of 300 gpm and is discharging t
Alchen [17]

Answer:

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Explanation:

8 0
3 years ago
Three samples of the same metal are weighed and their masses are found to be 44.40 g, 40.58 g, and 38.35 g. The corresponding vo
lidiya [134]

The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are, \frac{44.40}{4.8} = 9.25g/mL, \frac{40.58}{4.7} = 8.634g/mL and \frac{38.35}{4.2} = 9.130g/mL respectively.

The equation of the standard deviation is √{∑(x  - \frac{}{x})÷N}

Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.

The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.

The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.

The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.

The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be \sqrt{0.070}=0.265. Thus the standard deviation of the density value is 0.265g/mL.  

5 0
3 years ago
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