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Vlad [161]
3 years ago
15

Using the 5M NaCI and 10% glucose stock solutions you generated, how would you prepare 100 ml of a solution that is both 150mM N

aCl and 1 % glucose?
Chemistry
1 answer:
fredd [130]3 years ago
4 0

Answer:

Take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

Explanation:

  • In order to calculate the required volume of the 5 M NaCl solution, we calculated the moles contained in a 100 mL solution that has a concentration of 150 mM:

0.1 L * 0.150 M = 0.015 moles of NaCl

With those moles we can calculated the required volume, using the concentration of the stock solution:

0.015 mol / 5 M = 0.003 L = 3 mL.

  • To make a solution that has a 1 % concentration of glucose, from a 10 % glucose solution, is the same as to make it ten times less concentrated. Thus, with a final volume of 100 mL, you would need to take 10 mL of the 10% glucose solution, because 100mL * 10/100 = 10.

So in order to prepare the solution, you would need to take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

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The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

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