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Veseljchak [2.6K]
3 years ago
14

Provide a method for separating the mixtures of two or three compounds, dissolved in a solvent diethyl ether. In each case one o

f the components will be a neutral compound. You should give your answer in the form of a flow chart (see Section 12.12). a. Benzophenone and tributylamine b. 4-Bromoaniline, 3-nitrobenzoic acid, and benzoin c. Fluorenone, octanoic acid, and dicyclohexylamine d. 1-Hexanol and 4-bromoaniline
Chemistry
1 answer:
lys-0071 [83]3 years ago
5 0

Answer:

The complete aerobic oxidation of glucose is coupled to the synthesis of as many as 36 molecules of ATP

Explanation:

Glycolysis, the initial stage of glucose metabolism, takes place in the cytosol and does not involve molecular O2. It produces a small amount of ATP and the three-carbon compound pyruvate. In aerobic cells, pyruvate formed in glycolysis is transported into the mitochondria, where it is oxidized by O2 to CO2. Via chemiosmotic coupling, the oxidation of pyruvate in the mitochondria generates the bulk of the ATP produced during the conversion of glucose to CO2. The biochemical pathways that oxidize glucose and fatty acids to CO2 and H2O.

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Which element has atoms that can bond with each other to form ring, chain, and network structures?
matrenka [14]
This is basics of organic chemistry - the element that has atoms that can bond with each other to form ring, chain, and network structures is (3) carbon.
The other elements available above cannot form such structures, whereas carbon, which is abundant in nature, can.
6 0
3 years ago
Read 2 more answers
3. A sample of carbon has a mass of
Sav [38]

- 34 {}^{15}

  • The power of -34 will be 15 because 15 zeroes are given....
8 0
3 years ago
One hundred million copper atoms are arranged in a line. Which is a reasonable estimate for the length of this line?
Anettt [7]
The size of an atom is roughly 10^-10 metres

10^-10 • 100,000,000 = 0.01 metres

Since 1 metre = 100 centimetre

0.01 x 100 = 1 centimetre

There’s your answer
8 0
3 years ago
What is the concentration in mass percent of a solution prepared from 50.0g nacl and 150.0g of water?
otez555 [7]
M(NaCl)=50.0 g
m(H₂O)=150.0 g

m(solution)=m(NaCl)+m(H₂O)

w(NaCl)=100m(NaCl)/m(solution)=100m(NaCl)/{m(NaCl)+m(H₂O)}

w(NaCl)=100*50.0/(50.0+150.0)=25%
3 0
4 years ago
Calculate the number of sodium ions, perchlorate ions, Cl atoms and O atoms in 17.8 g of sodium perchlorate. Enter your answers
Degger [83]

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

17.8 g \times \frac{1mol}{122.44g} = 0.145 mol

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

0.145 mol \times \frac{6.02 \times 10^{23}molecules  }{mol} = 8.73 \times 10^{22}molecules

NaClO₄ is a strong electrolyte that dissociates according to the following equation.

NaClO₄ ⇒ Na⁺ + ClO₄⁻

The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion Na^{+} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion Na^{+}

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion ClO_4^{-} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion ClO_4^{-}

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

8.73 \times 10^{22}ion ClO_4^{-} \times \frac{1 atomCl }{1ion ClO_4^{-}} = 8.73 \times 10^{22}atom Cl

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

8.73 \times 10^{22}ion ClO_4^{-} \times \frac{4 atomO }{1ion ClO_4^{-}} = 3.49 \times 10^{23}atom O

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

You can learn more Avogadro's number here: brainly.com/question/13302703

8 0
2 years ago
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