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3 years ago

Without kidneys, your body cannot _____.

1 answer:

5 0

Hey There!

Here is your answer:

Without your kidneys the body cannot get rid of all the wastes it has. The kidneys produce urine which takes all the waste out of your body.

Which means its can not fliter liquid waste! Hope this helps!

You might be interested in

How many significant figures are there in the measurement 64,000 mg?

Vsevolod [243]

I'm pretty sure that significant figures is just the amount of numbers there is. So, in this case I think the answer would be D. 5

4 0

3 years ago

What are the ion concentrations in a 0.12 M solution of AlCl3? A) 0.12 M Al3+ ions and 0.040 M Cl- ions B) none of the answers C

KIM [24]

Answer:

Explanation:

To answer this question, we will need to write the dissociation equation of aluminum trichloride.

AlCl3 ——-> Al3+ + 3Cl-

It can be seen that when aluminum chloride dissociates, it gives one mole of aluminum ion and three moles of the chloride ion.

From here we can see that the concentration of the aluminum chloride equals that of the aluminum ion while that of the chloride ion is thrice that of the aluminum chloride. This means we simply multiply 0.12M by 3 to get the molarity of the chloride ion while that of the aluminum ion remains the same

5 0

3 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

2 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$