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aev [14]
3 years ago
5

Some of the following salts are used in detergents and other cleaning materials because they produce basic solutions. Which of t

he following could not be used for this purpose? Write equations to justify your answers.
a. Sodium carbonate
b. Sodium sulfate
c. Ammonium sulfate
d. Sodium phosphate
Chemistry
1 answer:
Ne4ueva [31]3 years ago
3 0

Answer: Sodium sulfate and Ammonium sulfate could not be used for this purpose.

Explanation:

An acidic solution will show a pH in the range of 0 - 6.9. A basic solution show a pH in the range of 7.1 - 14. A neutral solution will show a pH of 7.

A salt is acidic, when it is formed from the combination of strong acid and weak base. A salt is basic, when it is formed from the combination of weak acid and strong base. A salt is neutral, when it is formed from the combination of weak acid and weak base or strong acid and strong base.

a. Sodium carbonate : is formed by the combination of strong base NaOH and a weak acid H_2CO_3[, thus forms a basic solution.

Na_2CO_3\rightarrow 2Na^++CO_3^{2-}

b. Sodium sulfate:  is formed by the combination of strong base NaOH and a strong acid H_2SO_4[, thus forms a neutral solution.

Na_2SO_4\rightarrow 2Na^++SO_4^{2-}

c. Ammonium sulfate:  is formed by the combination of weak base NH_4OH and a strong acid H_2SO_4[, thus forms a acidic solution.

(NH_4)_2SO_4\rightarrow 2NH_4^++SO_4^{2-}

d. Sodium phosphate:  is formed by the combination of strong base NaOH and a weak acid H_3PO_4[, thus forms a basic solution.

Na_3PO_4\rightarrow 3Na^++PO_4^{3-}

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What can crystalize size tell us about where an igneous rock formed?
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Answer:

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4 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

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3 years ago
In acetyl CoA formation, the carbon-containing compound from glycolysis is oxidized to produce acetyl CoA. From the following co
kondor19780726 [428]

Answer:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Explanation:

Hello,

In this case, it is important to recall that acetyl-CoA is produced either by oxidative decarboxylation of pyruvate derived from glycolysis, which is carried out into the mitochondrial matrix, by cause of the oxidation of high-order fatty acids, or by oxidative degradation of very specific amino acids. Acetyl-CoA then enters in the citric acid cycle where it is oxidized in the light of energy production.

In this manner, during such processes, there are some net inputs and outputs, therefore, they are sorted as show below, considering there some of them not classified neither as input nor output:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Best regards.

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