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Makovka662 [10]
3 years ago
9

How did Mendeleev's work help in the discovery of new elements ?

Chemistry
2 answers:
KengaRu [80]3 years ago
6 0
It helped unlock many types of door to science and life
BARSIC [14]3 years ago
5 0
It helped future scientists with their studies
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stearic acid and linoleic acid stearic acid and linoleic acid stearic acid has the higher melting point, because it has two more
elena55 [62]

Acid palmitic acid has higher melting point, because it has two more methylene groups.

\text { Myristic acid } \rightarrow \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{12} \mathrm{COOH} \\& \text { Palmitic acid } \rightarrow \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{14} \mathrm{COOH} \\

Acid palmitic acid has higher melting point, because it has two more methylene groups.

Giving it a greater surface area and therefore more intermolecular  van der waals interact than the myristic acid.

stearic arid \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{16} \mathrm{COOH}

linoleic acid \mathrm{C}_{18} \mathrm{H}_{32} \mathrm{O}_2 (two double bond)

Stearic acid has higher Melting point, because it does not have any Carbon-Carbon double bonds, whereas linoleic acid has two cis double bonds which prevent the molecules from packing closely together.

Oleic Acid and Linoleic acid.

\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_2-one double bond (cis)

Acid palmitic acid has higher melting point, because it has two more methylene groups.

For more such question on methylene group.

brainly.com/question/4279223

#SPJ4

4 0
1 year ago
Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,
Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: K_{c} = 2.10^{-2}

Explanation: The reaction for steam reforming methane is:

CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = \frac{20}{125} = 0.16M

<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

4 0
3 years ago
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