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Charra [1.4K]
3 years ago
12

A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes t

he glass at a 41.3∘ angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. Methanol has a the refractive index of 1.329.

Chemistry
1 answer:
anyanavicka [17]3 years ago
4 0

18.6

use snells law for glass n=sini/sinr

the value of angle of refraction in glass becomes value of angle of incidence in the glass methanol interface

use snells law again to get the angle light makes with methanol.

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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

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then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

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by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

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3 years ago
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Rom4ik [11]

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6 0
3 years ago
Read 2 more answers
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