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Alecsey [184]
3 years ago
15

Is there DNA in the stucture of prokryotic

Chemistry
2 answers:
azamat3 years ago
5 0

Answer:

i hope this helped

They have no true nucleus as the DNA is not contained within a membrane or separated from the rest of the cell, but is coiled up in a region of the cytoplasm called the nucleoid. Prokaryotic organisms have varying cell shapes.

Explanation:

Jet001 [13]3 years ago
3 0

Answer:

Explanation:

The DNA in prokaryotes is contained in a central area of the cell called the nucleoid, which is not surrounded by a nuclear membrane. Many prokaryotes also carry small, circular DNA molecules called plasmids, which are distinct from the chromosomal DNA and can provide genetic advantages in specific environments.

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Write the balanced equation for the ionization of the weak base pyridine, c5h5n , in water, h2o. Phases are optional.
Lelu [443]

The balanced equation for the ionization of the weak base pyridine,C5H5N in water, H2O

C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)

<h3>What is the balanced equation for the ionization?</h3>

Generally, Pyridine is characterized by a ring structure, in this characteristic ring structure N is sp2 hybridized, hence creating a lone pair present on N so s - character is more, as well as lone pair, is present.

Therefore, Considering The following functions of the equation:weak base pyridine,C5H5N in water, H2O

We write the balanced equation for the ionization as

C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)

Read more about  Chemical Reaction

brainly.com/question/11231920

6 0
2 years ago
what is the percent by mass of NaHCO3 in a solution containing 20.0 g of NaHCO3 dissolved in 600.0 ml of water
Ne4ueva [31]
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%

m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water

%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
7 0
3 years ago
5TH GRADE SCIENCE!!!!!!
Yuri [45]

Answer:

Phase changes that require a loss in energy are condensation and freezing.

Explanation:

4 0
3 years ago
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While in science class a student measured a cube. She
alexira [117]

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7cm

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3 years 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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