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2 years ago

Glucose and amino acids are reabsorbed from the glomerular filtrate by the __________.

Chemistry

1 answer:

AleksandrR [38]2 years ago

5 0

Answer:

Amino acids, along with glucose, are reabsorbed in the glomerular system with a passive or active mechanism as the fluid travels through the entire renal tubular system and enters the circulation again.

Active mechanisms are those that require expenditure of energy, that is, expenditure of the energy currency, while the passive ones do not, they occur through spontaneous non-energy processes such as osmosis, the osmotic gradient and the difference in concentrations in different compartments.

Explanation:

Glomerular filtration is the regulator of the excretion of metabolites and toxic molecules or not necessary for our body. That is why if the amino acid values are high as well as those of glucose in urine, we will be facing a pathology.

If glucose is increased, it is because there is a glycemic peak in blood volume, hence possible diabetes.

And if the amino acids are increased, we could be facing an autoimmune or proteolytic pathology where a large amount of body proteins such as muscle proteins would be breaking down and releasing the amino acids that make it up, this phenomenon usually appears in those people who suffer from rhabdomyolysis in expenses very intense energy sources not appropriate.

On the other hand, glomerular filtration occurs in the kidney and is carried out by the nephron, which is the functional unit of the kidney, within it there is a specific tubular system in collection, absorption and reabsorption, added to the presence of Bowman's capsule.

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Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

Complete text:

Honors Stoichiometry Activity WorksheetInstructions:

Activity Two: Just Lemons, Inc. Production

Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.

Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients.

Water 946.36 g

Sugar 196.86 g

Lemon Juice 193.37 g

Lemonade 2050.25g

Leftover Ingredients?

Just Lemons Lemonade Recipe Equation:

2 water + sugar + lemon juice = 4 lemonade

Mole conversion factors:

1 mole of water = 1 cup = 236.59 g

1 mole of sugar = 1 cup = 225 g

1 mole of lemon juice = 1 cup = 257.83 g

1 mole of lemonade = 1 cup = 719.42 g

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

Water:

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

Sugar:

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

Lemon juice:

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0

2 years ago

PLS HELP ASAP, WILL GIVE 100 POINTS.

Ivan

Answer: values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

location of the electron: In the 7th energy level away from the nucleus.

Explanation:

From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.

The orbital designation of the describe electron is 7d

Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus.

Got it from quasarJose

Hope it helps

5 0

2 years ago

Large quantities of tritium were released into the atmosphere in the 1950s and 1960s during atmospheric nuclear testing. This tr

Sergeu [11.5K]

The ocean
hope I helped

8 0

2 years ago

Read 2 more answers

If excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration

Ray Of Light [21]

Answer:

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

Data Given:

Ksp of CaSO4 = 2.4 × 10∧-5

CaSO4 ⇔ Ca+2 + SO4∧-2

Solution:

Ksp = [Ca+2].[ SO4∧-2]

2.4 × 10∧-5 = [x].[x]= x²

x = 4.9×10∧-3 M

Result:

The conc. of Ca+2 ion is 4.9×10∧-3 M

3 0

2 years ago

What mass of carbon dioxide could be made from 100 tonnes of calcium carbonate?

MA_775_DIABLO [31]

Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3

i.e 100 tonnes of CaCO3 .

mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2

mass % of CO2 in CaCO3 = (44/100)×100 =44%

i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.

therefore, 44% of CO2 is present in CaCO3.

3 0

2 years ago