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stepan [7]
2 years ago
9

The human eye has an osmotic pressure of 8. 00 atm at 37. 0 °c. what concentration (in moles/l) of a saline (nacl) solution will

provide an isotonic eyedrop solution?
Chemistry
1 answer:
Gelneren [198K]2 years ago
7 0

0.3147 concentration (in moles/l) of a saline (NaCl) solution will provide an isotonic eyedrop solution.

Isotonic eye drops

Because it might result in eye discomfort or tissue damage if it is not maintained, isotonicity is regarded as a crucial component of ophthalmic medicines. A few drops of blood are mixed with the test preparation before being examined and judged under a microscope at a magnification of 40. Isotonic solutions are those that have the same amount of water and other solutes in them as the cytoplasm of a cell. Since there is no net gain or loss of water, placing cells in an isotonic solution will not cause them to either shrink or swell.

We can calculate the osmotic pressure exerted by a solution using the following expression.

π = M . R . T

where,

π is the osmotic pressure

M is the molar concentration of the solution

R is the ideal gas constant

T is the absolute temperature

The absolute temperature is 37 + 273 = 310 K

π = M . R . T

8 = (X mol/L) . (0.082atm.L/mol.K) . 310 K = 0.3147 mol/L

To learn more about osmotic pressure refer:

brainly.com/question/5041899

#SPJ4

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A solution of carbonic acid is at equilibrium. How would the system change is more carbonic acid was added to the solution
7nadin3 [17]

Answer:

The equilibrium position shifts to the right, in accordance to the constraint principle

3 0
3 years ago
199.5 grams unrefined dark crystalline sugar to cups
Dmitry [639]

Answer:

0.9975 cup  

Step-by-step explanation:

"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."

200.0 g brown sugar = 1 cup

 199.5 g brown sugar = 199.5× 1/200 .0

 199.5 g brown sugar = 0.9975 cup

A standard measuring cup is not capable of this precision and, furthermore, the mass of brown sugar you can get into a cup depends on how tightly you pack it.

Your Mole Day cake will be fine if you use 1 cup of brown sugar as usual.

4 0
3 years ago
In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

4 0
3 years ago
Please help i need this as soon as possible
UNO [17]

Answer:

1. 1, 3, 2

2. 2, 2, 3

3. 2, 1, 2, 1

4. 2, 1, 2

5 0
2 years ago
g The reaction C(s) + CO2(g) → 2CO(g) is spontaneous only at temperatures in excess of 1100 K. We can conclude that a. ΔG° is ne
mel-nik [20]

Answer:

c. ΔH° is positive and ΔS° is positive.

Explanation:

Hello,

In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

\Delta G=\Delta H-T\Delta S

Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.

Best regards.

7 0
3 years ago
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