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2 years ago

The following atoms are all variations of the element carbon: 12C, 13C, and 14C due to differing

1 answer:

5 0

Isotopes.
Some isotopes of elements that are not usually radioactive are radioactive because they don't have a suitable number of neutrons. For example carbon-12 is stable but carbon-14 is radioactive. Carbon-12 is much more common.

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After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage

Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have 2 reactions, the formation of the grignard reagent and the formation of the alcohol. The first step then is the calculation of the maximum amount of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the following conversion ratios:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide ( $C_6H_5Br$ )= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and $C_6H_5Br$ = 1:1

$2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg$

$9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br$

The smallest value is the mol of bromobenzene therefore 0.0402 mol of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of formation of the alcohol. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

$4.83~g~\frac{1~mol}{136.14~g}=0.35~mol$

The we have to divide by the coefficient of each reactive in the balance reaction. So:

$\frac{0.35~mol}{1}=0.35$

$\frac{0.0402~mol}{2}=0.0201$

Therefore the limiting reagent would be the phenylmagnesium bromide. Now, the molar ratio between the phenylmagnesium bromide and triphenyl carbinol is 2:1, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to grams of triphenyl carbinol:

Molar mass of triphenyl carbinol= 260.33 g/mol

$0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol$

Finally, we have to divide the obtanied solid by the calculated one:

$Percentage=\frac{5}{5.23}*100=95.6\%$

5 0

2 years ago

How is phosphorylation of glyceraldehyde 3-phosphate in the payment phase of glycolysis different from phosphorylation of glucos

svet-max [94.6K]

In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.

Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).

The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.

This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.

For more information on phosphorylation click on the link below:

brainly.com/question/7465103

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7 0

2 months ago

Dahasolnce [82]

Answer:

22.5moles

Explanation:

using mole ratio,n(CO2)/n(o)=1/2

n(CO2)=1/2n(02)

n(CO2)=1/2×45

moles of carbon dioxide produced is 22.5moles

8 0

2 years ago

13. Would you expect oxygen to form a cation or anion? How many electrons would it gain or lose? Why?

netineya [11]

Answer: C)Anion, it would gain 2 electrons to satisfy the octet rule.

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electrons are filled according to Afbau's rule in order of increasing energies and thus the electronic configuration of oxygen with 8 electrons is

$O:8:1s^22s^22p^4$