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zzz [600]
3 years ago
12

A 0.500 m solution of a weak base has a ph of 9.75. what is the base hydrolysis constant, kb, for the weak base?

Chemistry
1 answer:
Vladimir [108]3 years ago
8 0
When we have PH = 9.75
So we can get POH = 14 - 9.75 = 4.25 
and when POH = - ㏒[OH-]
by substitution:
4.25 = -㏒[OH-]
∴[OH] = 5.6x10^-5
from this reaction equation:
BOH ↔ B+  + OH-
∴[OH-] = [B+]= 5.6x10^-5 M
and Equ [BOH] = 0.5 m - X
                          = 0.5 - (5.6x10^-5)
                          =  0.4999
∴ Kb = [OH-][B+]/[BOH]
         = (5.6x10^-5)^2 / 0.4999
         = 6.27x10^-9

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Answer:

yes  it can  λ =265  nm

Explanation:

Here we will use  the relationship

   E = h c/λ  ∴   λ  = E/ hc  where

                                          h= Plank's constant

                                           c= Speed of light

                                           λ = Wavelength = ?

Substituting

note need E in J ,

E = 4.7 eV x 1.602 x 10⁻¹⁹ J/eV = 7.5 x 10⁻¹⁹ J)

λ = 7.5 x 10 ⁻¹⁹ J / (  6.626 x 10⁻³⁴ Js x 3 x 10^8) =  2.65 x 10⁻⁷ m = 2.65

=  2.65 x 10⁻⁷ m x 1 x 10⁹ nm/m = 265 nm

4 0
3 years ago
Describe the differences among primitive, igneous, sedimentary, and metamorphic rock, and relate these differences to their orig
Akimi4 [234]

Answer:

Rocks are the aggregate of minerals. There are three distinct categories of rocks, namely the sedimentary, metamorphic and the igneous rocks.

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  • The metamorphic rocks are derived from the previously existing sedimentary, igneous or other metamorphic rocks, due to the influence of extremely high pressure as well as temperature conditions. For example, Quartzite and Marble.
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All these three types of rocks are formed by different processes and their mode of origins are also different.

5 0
3 years ago
For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con
Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

A\rightarrow B

Expt    [A] M        [B] M         Rate [M/s]

1          3.40         4.16           1.82*10^-4

2         4.59         4.16           3.32*10^-4

3.        3.40          5.46          1.82*10^-4

Rate = k[A]^{x}[B]^{y}

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2

To find y:

Divide rate of expt 3 by expt 1

\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0

Therefore: x = 2, y = 0

Rate = k[A]^{2}[B]^{0}

To find k

Use rate for expt 1:

k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1

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