All categories

1 year ago

If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?

Chemistry

2 answers:

expeople1 [14]1 year ago

4 0

Answer: The cell will shrink when kept in 1% NaCl solution.

Explanation: This is explained with the help of osmosis.

Osmosis is a process in which the solvent flow from a solution of low concentration to a solution of high concentration through a semi-permeable membrane.

We a given a cell which is isotonic to 0.88% NaCl solution.

When the same cell is kept in a solution of 1% NaCl solution, osmosis will occur from inside to outside because the concentration is low inside the cell than the outside. So, the solvent will flow from inside to outside, which results in the shrinkage of cell. The solutions which have high concentration outside the cell are known as hypertonic solutions.

EleoNora [17]1 year ago

3 0

<span>The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to </span>extracellular fluid and equilibrium is reached.

You might be interested in

Two glasses labeled A and B contain equal amounts of water at different temperatures. Kim put an antacid tablet into each of the

skelet666 [1.2K]

Answer:

Option D. The water in Glass A is cooler than the water in Glass B; therefore, the particles in Glass A move slower.

Explanation:

Solubilities of solutes are enhanced when the temperature is increased.

From the experiment conducted,

It is evident that glass B temperature is higher than glass A temperature, because the solute dissolves faster in glass B than in glass A . This implies that glass A is cooler than glass B, hence the particles in A will move slower than that in B.

6 0

2 years ago

Read 2 more answers

An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

jekas [21]

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

For example, $O_{2}$ , $N_{2}$ etc are pure substances.

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

pure chemical substance.

heterogenous mixture.

mixture.

4 0

1 year ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

2 years ago

How many grams of methane gas (CH4) occupy a volume of 11.2 liters at STP

Vsevolod [243]

11.2L/22.4L (STP value) x 1 mol of CH4 x 16.04 g of CH4 = 8.2 g

7 0

2 years ago

Read 2 more answers

Which statement best describes how electrons fill orbitals in the periodic table? Electrons fill orbitals in order of their incr

Stolb23 [73]

Electrons fill orbitals in order of increasing energy from left to right. As the group number increase also the number of valence electorns of each group will increases

5 0

2 years ago

Read 2 more answers