for a concentration cell, the cell potential increases as: select the correct answer below:
A. the discrepancy in concentration between the half cells
B. increases the discrepancy in concentration between the half cells
C. decreases the total volume of the half cells
D. increases the total volume of the half cells

Answer:

It would increase.

Explanation:

Answer:

All human beings are 99.9 percent identical in their genetic makeup. Differences in the remaining 0.1 percent hold important clues about the causes of diseases

Explanation:

At pH 6, the dominant species present in water for aniline (C6H7N) is the deprotonated form, known as the anilide ion (C6H6N-).

To determine the dominant species of aniline (C6H7N) in water at a constant pH of 6, we need to compare the pKa (acid dissociation constant) of aniline with the pH. The pKa of aniline can be calculated using the pKb value provided. The relationship between pKa and pKb for a given species is as follows:

pKa + pKb = 14

Rearranging the equation, we find:

pKa = 14 - pKb

Therefore, the pKa of aniline is: pKa = 14 - 9.4 = 4.6

Now, comparing the pKa to the pH of 6, we can determine the dominant species present.

When the pH is lower than the pKa (pH < pKa), the acidic form predominates.

When the pH is higher than the pKa (pH > pKa), the basic form predominates.

When the pH is close to the pKa (pH ≈ pKa), both the acidic and basic forms are present in similar concentrations, and the solution is a buffer.

In this case, the pH of 6 is higher than the pKa of aniline (pH > pKa), so the basic form predominates.

Therefore, at pH 6, the dominant species present in water for aniline (C6H7N) is the deprotonated form, known as the anilide ion (C6H6N-).

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Answer:

Esophagus

Explanation:

Answer:

D. External sex organs form .

Explanation:

There are three stages in fetal development that include germinal, embryonic, and fetal period.

Three months means, it will be the 12 week of the fetal development and called fetal period. At the end of this stage, baby's external genitalia or sex organs will start developing such as penis in male and vagina in female.

Hence, the correct option is D.

Answer:

Vocal cords develop.

Explanation:

Explanation:

if solute is 43% inside the cell then water is 57% (100%-43% = 57%)

outside; 100% - 60% = 40% water

it is hypertonic because water concentration is high inside the cell than outside, hence water flows to the outside of the cell.

the momentum of the ball is 270.9

Answer:

Try uh hydrolysis I guess. If it's not, it's photosynthesis?

I'm sorry this is useless, I will research it, I'm sorry.

Answer:

Explanation:

-how continents match up when put together like a puzzle

-fossils of the same animals found in two different continents

The following ways did the Green Revolution increase food production?

The green revolution is a global agricultural system reform that took place from the 1950s to the 1960s. The green revolution aims to increase food productivity by transforming traditional agriculture into modern agriculture.

The green revolution led to high crop productivity through adapted measures, such as

This question is multiple choice:

The correct answer is E

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Answer:

The purpose of repeating a simulation is to ensure the validity of the results and fix any mistakes. The more times you repeat a simulation the more accurate the results and data will be.

Hope this helps!

- Quinn <3

I don't think so much for all your friends and family

Answer:

Explanation:

1.childhood

2.greenest

3.funny

Answer: established bluegrass, fescue, bentgrass, Bermudagrass, bahiagrass, centipedegrass, zoysiagrass, ryegrass

Explanation:

Answer:

melanin production

Explanation:

it occurs in the hair or skin

The type of intercellular junction that functions as a rivet or "spot weld" is a(n) Desmosomes.

Animal cells may additionally comprise junctions referred to as desmosomes, which act like spot welds among adjoining epithelial cells. Intercellular matrix can be liquid, gel or rigid. Skeletal, cardiac and smooth. A sort of connective tissue that shops fat. Contains difficult mineral containing intercellular matrix, consists of osteocytes and is a sort of connective tissue. Lining epithelium: The urinary bladder lining is a specialised stratified epithelium, the urothelium. The urothelium is solely in urinary systems which include the ureter, urinary bladder, and proximal urethra.

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The expected rate of return of Freedman Flowers is 9.9%.

Expected rate of return is the sum of the multiplication of each possible return with its corresponding probability of occurrence. Freedman Flowers' stock has a 50% chance of producing a 25% return, a 30% chance of producing a 10% return, and a 20% chance of producing a -28% return. The expected rate of return of Freedman Flowers can be calculated as follows:

Expected rate of return = (Probability of return 1 × Return 1) + (Probability of return 2 × Return 2) + (Probability of return 3 × Return 3)

Therefore, the expected rate of return of Freedman Flowers is:

Expected rate of return = (0.5 × 25%) + (0.3 × 10%) + (0.2 × -28%)

Expected rate of return = 12.5% + 3% - 5.6%

Expected rate of return = 9.9%

Therefore, the expected rate of return of Freedman Flowers is 9.9%.Option B, 9.9006 is the closest to the calculated expected rate of return.

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The rate of change of the volume of the balloon (sphere) is 150π cubic cm/min. The formula for the volume of a sphere is given by; V = (4/3)πr³.

Given that the circumference of a balloon (sphere) is increasing at a rate of 3π cm/min and the radius is 5 cm, we have to find the rate of change of the volume of the balloon. We know that the formula for the volume of a sphere is given by; V = (4/3)πr³. Differentiating with respect to time gives; dV/dt = 4πr² × dr/dt

We know that the circumference of a sphere is given by; C = 2πr

Differentiating with respect to time gives; dC/dt = 2π × dr/dt

Now we can find the value of dr/dt from the above expression; dr/dt = dC/dt ÷ 2πPutting the given value in the above expression, we have ;dr/dt = 3π ÷ 2πdr/dt = 3/2 cm/min

Now substituting the value of r and dr/dt in the expression for dV/dt; dV/dt = 4πr² × dr/dt dV/dt = 4π(5)² × (3/2) dV/dt = 150π

Therefore, the rate of change of the volume of the balloon (sphere) is 150π cubic cm/min.

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Based on the provided characteristics, the isolate from the cervical mucosal abscess specimen would most likely be Porphyromonas species.

The key features that point to this conclusion are:

Anaerobic gram-negative bacillus: Porphyromonas species are anaerobic bacteria and appear as gram-negative bacilli.

Inhibited by bile: Porphyromonas species are known to be sensitive to bile.

Production of black pigment: Porphyromonas species are known to produce a black pigment.

Resistant to vancomycin: Porphyromonas species are generally resistant to vancomycin.

Negative for indole production: This characteristic is consistent with Porphyromonas species.

Positive for glucose, sucrose, and lactose fermentation: Porphyromonas species typically ferment glucose, sucrose, and lactose.

Taken together, these characteristics align with the identification of Porphyromonas species. It's important to note that further confirmation and identification may be required through additional laboratory testing and analysis to definitively determine the specific species within the Porphyromonas genus.

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The population of frogs in a pond grows exponentially, tripling every 6 days. Initially, there are 28 frogs.

Time to 260 frogs:

A = P(1 + r/n)^(nt)

260 = 28(1 + 3/1)^(6t)

9.2857 = 4^(6t)

(6t) log 4 = log 9.285

t = log 9.2857 / log 4

2.378 days to reach 260 frogs.

Time to 1300 frogs:

A = P(1 + r/n)^(nt)

1300 = 28(1 + 3/1)^(6t)

46.4286 = 4^(6t)

(6t) log 4 = log 46.4286

t = log 46.4286 / log 4

6.027 days until the pond's ecosystem can support 1300 frogs.

Therefore, it will take around 2.378 days to reach 260 frogs and approximately 6.027 days until the situation becomes critical when the pond's ecosystem can support 1300 frogs.

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In this ecological coral reef study, we used a control impact approach, BACI.

Unfortunately, coral reefs suffer from a number of threats, ranging from

Furthermore, climate change further exacerbates these local threats.

A species of coral reef fish that is threatened by both overfishing and pollution. the observational study that we can use to understand the relative importance of these two threats would be observation over a certain period of time. this study would deal with any potentially confounding variables using reference articles.

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The enzyme responsible for building a new copy of DNA by adding the proper nucleotides to the template DNA strand is DNA polymerase.  DNA polymerase is and how it works.DNA Polymerase DNA polymerase is an enzyme that plays an essential role in the replication of DNA.

It is responsible for the synthesis of new strands of DNA by adding nucleotides to the existing template strand in the 5' to 3' direction. DNA polymerase requires a primer, which is a short sequence of RNA that is complementary to the template strand and provides a free 3'-OH group for nucleotide addition.DNA polymerase has a proofreading function that helps maintain the accuracy of DNA replication. As it synthesizes new strands of DNA, it checks to make sure that each nucleotide is correctly base-paired with the complementary nucleotide on the template strand.

If a mistake is made, the DNA polymerase will backtrack and remove the incorrect nucleotide before continuing with synthesis.In summary, DNA polymerase is an enzyme that adds the proper nucleotides to the template DNA strand during DNA replication. It plays an essential role in the accurate and efficient duplication of genetic material.

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When a red blood cell is placed in a hypertonic solution, it shrinks or undergoes crenation due to water loss through osmosis.

When a red blood cell is placed in a hypertonic solution, it means that the concentration of solutes in the solution is higher than the concentration of solutes in the intracellular fluid. As a result, water will move out of the cell through osmosis, following the concentration gradient. The hypertonic solution causes the cell to shrink or undergo crenation. This occurs because the higher solute concentration in the external solution draws water out of the cell, leading to a decrease in cell volume.

The hypertonic environment creates an osmotic imbalance, where water leaves the cell to equalize the concentration of solutes inside and outside the cell. The cell membrane contracts inward, causing the cell to become smaller and wrinkled. In the case of red blood cells, crenation can impair their ability to carry oxygen and nutrients and can also disrupt their normal function in the bloodstream. Overall, the hypertonic solution causes the red blood cell to undergo a process of dehydration and shrinkage.

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Answer:

With an updraft, downdraft, and rain, the cloud is now called a cumulonimbus cloud and the cycling of air up and down is called a thunderstorm cell. The moving air within the cloud builds up electric charges as it slides past other air.

Explanation:

^^'''

Answer: Water droplets freeze in the cumulonimbus clouds.

Explanation: I took the test on edge 2022.

The macromolecule pictured above is Fat. Option B is correct.

Fats are composed of glycerol, a type of monomer, and fatty acid, another type of monomer. Fat acts as an insulator, a cushion, and a permanent energy source for living things. Fats are also the building blocks of cell membranes and are used to store vitamins, filter toxins, and synthesize hormones.

Types of lipids include triglyceride, phospholipid, and sterol. In this article, we’ll look at the structure of these three lipids, how they work in the body, and where they are found in food.

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Answer: A change in the sequence of amino acids determined by the gene

Regulating gene expression is a key process in both prokaryotic and eukaryotic cells. Controlling how an RNA transcript is splicedis NOT used in prokaryotic cells.

Here, correct option is D.

In prokaryotic cells, gene expression is regulated by controlling which mRNAs get translated into proteins by the ribosomes, controlling how often a gene is transcribed, and controlling how rapidly proteins are destroyed once they are made. In eukaryotic cells, these are all regulatory mechanisms used as well, but there are additional methods of regulating gene expression that are not found in prokaryotes.

These include controlling how an RNA transcript is spliced and regulating the rate of transcription through the use of transcription factors. Splicing involves the removal of introns from a pre-mRNA molecule, resulting in a mature mRNA molecule that can be translated into a functional protein.

Therefore, correct option is D.

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Answer:

In both animal and plant cells. ... Only in plant cells. Mitochondria. Produces a usable form of energy for the cell.

The main answer to the question is d) XPD.

XPD is a subunit of TFIIH and mutations in this gene can cause the genetic disease xeroderma pigmentosum.

An explanation for this is that TFIIH is a multi-subunit complex that is involved in transcription initiation and DNA repair.

XPD is a DNA helicase that is a critical component of TFIIH and is responsible for unwinding DNA during transcription and DNA repair.

Mutations in the XPD gene can disrupt TFIIH function, leading to defects in DNA repair and increased susceptibility to UV-induced skin cancers and other symptoms associated with xeroderma pigmentosum.

In summary, XPD is a subunit of TFIIH that is involved in DNA repair and mutations in this gene can cause xeroderma pigmentosum.

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The given statement that bacteria are about five to ten times larger than yeasts and protozoa is false. In reality, bacteria are much smaller in size than yeasts and protozoa.

Bacteria are a type of unicellular microorganisms that belong to the prokaryotic group. They are the simplest and most abundant living organisms on earth, and they can be found in almost every environment, including water, soil, air, and the human body. Bacteria are incredibly small in size, ranging from about 0.2 to 10 micrometers (μm) in length. They are so small that they cannot be seen with the  and can only be viewed under a microscope. Yeasts are a type of unicellular fungi that are larger than bacteria. They are eukaryotic organisms that can be found in various habitats, including soil, water, and plant surfaces. Yeasts range in size from about 3 to 40 μm in length, which is much larger than the size of bacteria. Protozoa are unicellular eukaryotic microorganisms that can be found in various aquatic and terrestrial environments. They are much larger in size than both bacteria and yeasts, ranging from about 5 to 500 μm in length. Protozoa are classified into different groups based on their locomotion, feeding, and reproduction methods. Therefore, the given statement that bacteria are about five to ten times larger than yeasts and protozoa is false, and the actual size order from smallest to largest is bacteria < yeasts < protozoa.

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