Devise a detailed experimental procedure to purify ~ 20 grams of benzoic acid that is contaminated with sodium chloride. Justification of the steps (including solubility calculations) that are included in the procedure. In other words, explain why the steps are being included.

Answers

Answer 1

Answer:

Answer:

Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.

Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.

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Indicate the number of protons, neutrons, and electrons in each of the following species: 15N7, 33S16, 63Cu29, 84Sr38, 130Ba56, 186W74, 202Hg80

Answers

The number of protons neutrons, and electrons in each of the following species given are below;

What is the atomic number?

The total number of protons present in an atom is known as the atomic number of that atom. The atomic number has no correlation either with the number of neutrons or the number of electrons present inside an atom.

15N7 ⇒ 7 electrons, 8 neutrons, 7 protons

33S16 ⇒ 16 protons, 16 electrons, 17 neutrons

63Cu29 ⇒ 29 electrons, 34 neutrons,29 protons

84Sr38 ⇒ 38 electrons, 46 neutrons,38 protons

130Ba56 ⇒ 56 electrons, 74 neutrons,56 protons

186W74⇒ 74 electrons, 112 neutrons,74 protons

202Hg80 ⇒ 80 electrons, 122 neutrons ,80 protons

To learn more about the atomic number from here, refer to the link ;

brainly.com/question/14190064

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Answer:

1. 7 protons, 7 electrons, 8 neutrons

2. 16 protons, 16 electrons, 17 neutrons

3. 29 protons, 29 electrons, 34 neutrons

4. 38 protons, 38 electrons, 46 neutrons

5. 56 protons, 56 electrons, 74 neutrons

6. 74 protons, 74 electrons, 112 neutrons

7. 80 protons, 80 electrons, 122 neutrons

A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture

Answers

There is 65% of NaHCO3 in the sample.

The equation of the reaction is;

HA + NaHCO3 -----> NaA + CO2 + H2O

Amount of CO2 formed = mass/molar mass

mass of CO2 = 0.561 g/44 g/mol = 0.013 moles

From the balanced reaction equation;

1 mole of NaHCO3 yields 1 mole of CO2

0.013 moles of Na2CO3 yields 0.013 moles of CO2

Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3

Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1

= 65%

Learn more: brainly.com/question/25150590

Answer:

63.75%.

Explanation:

The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;

HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).

Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.

Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.

Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.

Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.

Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.

4. Oxygenating myoglobin. The myoglobin content of some human muscles is about 8 g kg 1. In sperm whale, the myoglobin content of muscle is about 80 g kg 1 . (a) How much O 2 is bound to myoglobin in human muscle and in sperm whale muscle? Assume that the myoglobin is saturated with O 2, and that the molecular weights of human and sperm whale myoglobin are the same. Berg, Jeremy M.. Biochemistry (p. 213). W. H. Freeman. Kindle Edition.