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miss Akunina [59]
2 years ago
11

The next 3 questions will walk you through using the Henderson-Hasselbalch equation for the following question. For each step pr

ovide a numerical answer. A solution is made by combining 300mL of 0.020M trimethyl amine, that is (CH3)3N and a 500mL of a 0.03M solution of it's conjugate acid, that is (CH3)3NH . The Kb for trimethyl amine is 6.3x10-5. The Henderson-Hasselbalch equation uses pKa not Kb, Calculate the pKa....
Chemistry
1 answer:
kicyunya [14]2 years ago
8 0

Answer:

<em>The pKa is 13.0.</em>

Explanation:

pKa + pKb = 14

Given, Kb of trimethylamine = 6.3 × 10^{-5}

pKb = - log (6.3 × 10^{-5})

= 1.0

⇒ pKa = 14 - pKb = 14 - 1.0

<u>pKa = 13.0</u>

<em><u></u></em>

<em>Check: For most weak acids,  pKa ranges from 2 to 13.</em>

You might be interested in
If a concentration of 10 fluorescent molecules per μm2of cell membrane is needed to visualize a cell under the microscope, how m
lara [203]

Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.

Explanation : As a single egg cell has an approximately diameter of 100 μm.

We can use this formula to calculate area of the cell membrane;

A = π (100)^{2} / 4;  

We can take π as 3.14 and we get;

A = 3.14 X (100)^{2} / 4  

Soving we get;

A =  7850 μm^{2}  

Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μm^{2}  but;

here 1 μm^{2} = 7850 μm^{2} dye molecules.

Therefore, 10 fluorescent molecules will need;  

7850 X 10 = 78500 molecules of dye.

Therefore, the answer is 78500 molecules of dye.

6 0
2 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
2 years ago
The volume of hcl gas required to react with excess ca to produce 11.4 l of hydrogen gas at 1.62 atm and 62.0 °c is ________ l.
seropon [69]

Answer:

22.8 L  

Step-by-step explanation:

We can use <em>Gay-Lussac's Law of Combining Volumes</em> to solve this problem:

Gases <em>at the same temperature and pressure</em> react in simple whole-number ratios.

1. Write the chemical equation.

Ratio:                 2 L                             1 L

          Ca(s) + 2HCl(g) ⟶ CaCl₂(s) + H₂(g)

V/L:                                                     11.4

2. Calculate the volume of HCl.

According to the law, 2 L of HCl form 1 L of H₂.

Then, the conversion factor is (2 L HCl/1 L H₂).

Volume of HCl = 11.4 L H₂ × (2 L HCl/1 L H₂)

                         = 22.8 L HCl

3 0
2 years ago
How many moles of PBr3 contain 3.68 x 10^25 bromine atoms?
scoray [572]
<span>3.68 x 10²⁵ bromine atoms * 1mol/6.02*10²³ atoms=
 = 61.13 mol of bromine atoms

1 mol PBr3 ----- 3 mol Br
x mol PBr3 -----61.13 mol Br

x= 1*61.13/3 = 20.4 mol PBr3.


</span>20.4 mol PBr3 <span>contain 3.68 x 10^25 bromine atoms.</span>
7 0
2 years ago
Read 2 more answers
A sample of TNT, C7H5N3O6 , has 8.94 × 1021 nitrogen atoms. How many hydrogen atoms are there in this sample of TNT?
Bess [88]

Answer:

1.49×10²² atoms of H are contained in the sample

Explanation:

TNT → C₇H₅N₃O₆

1 mol of this has 7 moles of C, 5 moles of H, 3 moles of N and 6 moles of O

Let's determine the mass of TNT.

Molar mass is = 227 g/mol

As 1 mol has (6.02×10²³) NA atoms, how many moles are 8.94×10²¹ atoms.

8.94×10²¹ atoms / NA = 0.0148 moles

So this would be the rule of three to determine the mass of TNT

3 moles of N are in 227 g of compound

0.0148 moles of N are contained in (0.0148 .227) / 3 = 1.12 g

Now we can work with the hydrogen.

227 grams of TNT contain 5 moles of H

1.12 grams of TNT would contain (1.12 .5) / 227 = 0.0247 moles

Finally let's convert this moles to atoms:

0.0247 mol . 6.02×10²³ atoms / 1 mol = 1.49×10²² atoms

8 0
2 years ago
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