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castortr0y [4]
4 years ago
14

Propose structures that meet the following descriptions. Use bond-line notation to draw your structural formulas. a. ketone with

five carbons and one alcohol c. alkane (no double bonds) with the formula, C6H12 e. nitrile with the formula, C5H9N g. phenol with a carboxylic acid group adjacent to hydroxyl i. amide with the formula, C3H7NO k. a secondary thiol with 3 carbons m. a primary alcohol with five carbons b. a-ketoester (i.e., a molecule with a ketone one carbon away from an ester) d. dialdehyde with the formula, C4H6O2 f. b-ketocarboxylic acid (i.e. a molecule with a ketone two carbons away from a carboxylic acid) h. amino acid (i.e. a carbon atom attached to both a carboxylic acid and an amine j. ether with the formula, C4H10O l. alkane with five carbons, one of which is a quaternary carbon n. hydroxy aldehyde (a compound containing an aldehyde and hydroxy group)

Chemistry
1 answer:
bija089 [108]4 years ago
7 0

Answer:

See explanation

Explanation:

In putting up prospective structures for the compounds, we must consider the tetra valency of carbon. Again you must read the description carefully before drawing each structure.

In each of the proposed structures, the number of atoms present must reflect the condensed structural formula given in the question.

Proposed structures for each chemical specie in the question are shown in the images attached.

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Which arrow or arrows represent reactions that demonstrate a conservation of mass and energy? Explain your answer
rewona [7]
I think it would be the sun because it’s the one giving energy to most things, and if you haven’t answered it yet you can put the sun because it’s giving energy to the plans and the environment and it’s the the sun was the effect of the whole thing, sorry if I didn’t help
6 0
3 years ago
The decay of radioactive elements occurs at a fixed rate. The half-life of a radioisotope is the time required for one half of t
Stolb23 [73]

Answer:

100 grams of C-14 decays to 25 grams in 11,460 years.

The C-14 isotope is only useful for dating fossils up to about 50,000 years old

If an ancient bone contains 6.25% of its original carbon, then the bone must be 22,920 years old.

Explanation:

We already know that the half life of C-14 is 5,730 years. After the first half life, we have 50 grams remaining. This takes 5,730 years. After the second half life (11,460 years now gone) we have 25 grams of C-14  left.

If a fossil material is  older than 50,000 years an undetectable amount of 14C is left in the sample hence Carbon-14 is no longer suitable for dating the sample.

From;

0.693/5730 = 2.303/t log (No/0.0625No)

Where;

t = time taken and No = initial amount of C-14

0.693/5730= 2.77/t

t = 22,920 years

6 0
3 years ago
Calculate the number of joules of heat energy needed to increase the temperature of 25.0 g of metal from 21.0 ºC to 80.0 ºC. The
Harman [31]

Answer:

Q = 768.47 J

Explanation:

Given that,

Mass of the metal, m = 25 g

Initial temperature, T₁ = 21.0 ºC

Final temperature, T₂ = 80.0 ºC

The specific heat of the metal is 0.521 J/gºC.

We know that the heat released due to the change in temperature is given by :

Q=mc\Delta T\\\\=25\times 0.521\times (80-21)\\Q=768.47\ J

Hence, 768.47 J of heat energy will be needed.

7 0
3 years ago
) an atom has 6 electrons in its outer shell. how many unpaired electrons does it have?
olasank [31]
It would be none would it not
8 0
3 years ago
During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).
Alexus [3.1K]

Answer:

The answers to the question are as follows

First part

The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g

Second part

The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F  

Explanation:

To solve this, we write the equation for the discharge of the lead acid battery as

H₂SO₄ → H⁺ + HSO₄⁻

Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻

at the cathode we have

PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O

Summing the two equation or the total equation for discharge is

Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

From the above one mole of lead and one mole of PbO₂  are consumed simultaneously hence

Number of moles of lead contained in 405 g of Pb with molar mass  = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb

Hence number of moles of  PbO₂ reduced at the cathode = 1.95 mole

mass of  PbO₂ reduced at the cathode = (number of moles)×(molar mass)

= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode

Part B

Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred

Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs

hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge

or transferred electrical charge = 377186.86 C or 3.909 Faraday

6 0
3 years ago
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