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3 years ago

How did agre use a simple osmosis experiment to prove the function of aquaporin?

1 answer:

6 0

Agre tested his hypothesis in a simple experiment where he compared cells which contained the protein in question with cells which did not have it. The trials were ran with artificial cells, liposomes, which are a type of soap bubble. He found that the liposomes became permeable if the protein was planted inside of the membrane.

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A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ),

Semenov [28]

Answer:

[KOH] : 1.47 M

[KOH] : 1.22 m

[KOH]: 6.42 % mass percent.

Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

7 0

3 years ago

In a redox reaction, reduction is defined by the:

jonny [76]

Answer:

C. Gain in electron(s) resulting in a decrease of oxidation number.

Explanation:

Redox reactions are reactions involving transfer of of electron between two species (reduction specie) and (oxidation species) and change resulting in change in oxidation number.

Reduction in terms of redox reaction is the specie that accepts electron(s) and gets "reduced" since its oxidation state has been reduced.

For example

Cl + e- → Cl⁻

The above reaction is an example of reduction reaction taking place in a redox reaction. We can see that Chlorine oxidation state was changed from (0) to (-1) state.

5 0

3 years ago

Explain the process of fossil fuel formation.

topjm [15]

Explanation:

Fossil fuel is an overall term for covered ignitable geologic stores of natural materials, framed from rotted plants and creatures that have been changed over to unrefined petroleum, coal, flammable gas, or weighty oils by introduction to warmth and weight in the world's outside more than a huge number of years.

The consuming of petroleum products by people is the biggest wellspring of emanations of carbon dioxide, which is one of the ozone depleting substances that permits radiative compelling and adds to an unnatural weather change.

A little bit of hydrocarbon-based powers are biofuels gotten from climatic carbon dioxide, and consequently don't build the net measure of carbon dioxide in the environment.

6 0

3 years ago

What is the percent composition by mass of oxygen in

KiRa [710]

What is the percent composition by mass of oxygen in magnesium oxide, MgO?

Answer: 39.7 percent

4 0

2 years ago

In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i

fgiga [73]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

7 0

3 years ago