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2 years ago

The standard hydrogen electrode:

Chemistry

1 answer:

tiny-mole [99]2 years ago

3 0

Answer:

is a simple, safe, and easy-to-create electrode

Explanation:

the hydrogen electrode is based on the redox half-cell:

2H+(aq) + 2e- → H2(g)

building:

1) Platinium electrode

2) hydrogen pumping

3) acid solution [H+] = 1M

4) siphon to prevent oxygen presence

5) galvanic cell connector

this electrode is used as the basis for standard potential tables

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Consider the element in the periodic table that is directly to the right of the element identified in part (a). Would the 1s pea

astraxan [27]

The question is incomplete. Here is the complete question.

The photoelectron spectroscopy is shwon below.

(a) Based on the photoelectron spectrum, identify the unknown element and write its electron configuration.

(b) Consider the element in the periodic table that is directly to the right of the element identified in part (a). Would the 1s peak of this element appear to the left of, the right of, or in the same position as the 1s peak of the element in part (a)? Explain your reasoning.

Answer and Explanation: Photoelectron Spectroscopy is a method of determinining the relative energy of electrons in atoms and molecules.

It is based on the photoelectric effect: when a radiation energy incides on a substance, an electron is ejected from it. If we know the kinetic energy of the ejected electron, known as photoelectrons, and the energy of the incident radiation, it is possible to find the energy of the electron in the substance.

The energy needed to eject an electron from the sample is called Binding Energy and in an atom, depends on which shell the electron is: valence eletrons (outermost shell), binding energy is lower; core eletrons (innermost shell), binding energy is highest.

In the graph, vertical axis shows 5 peaks for different energies. The peak closer to the origin, the leftmost peak, correspond to the 1s subshell, since their are closest to the nucleus, and so, has the highest binding energy.

Following from left to the right, we noticed:

First, second and fourth peaks has the same height;
Third peak's height is 3x higher than 1st, 2nd and 4th;
Fifth peak is one unit higher than first, second and fourth;

(a) Then, we can conclude the eletron configuration of the element is

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}$

which is Phosphorus with atomic number of 15.

(b) The element to the right of element P is Sulfur (S). The peak 1s of sulfur will appear in the same position as the 1s peak of Phosphorus, because the elements in the Periodic Table are grouped according to certain properties. Elements in the same horizontal line are elements in the same period, which one of the characteristics is they have the same total number of electron shells.

4 0

1 year ago

A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi

viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette. A blank had an absorbance of 0.029. The absorbance of an unknown solution of compound A was 0.375. Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$

Explanation:

According to the Lambert's Beer law:-

$A=\epsilon l c$

Where, A is the absorbance

l is the path length

$\epsilon$ is the molar absorptivity

c is the concentration.

Given that:-

c = $3.96\times 10^{-4}\ M$

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

$0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M$

$\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}$

Molar absorptivity of compound A = $1502.53\ {Ms}^{-1}$

4 0

2 years ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

2 years ago

The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =

Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) (1)

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] (2)

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) = 9.744g of monosodium succinate.

Mass of disodium succinate:

0.0304moles * (162g / 1mol) = 4.925g of disodium succinate.

5 0

2 years ago

30. how many grams of boric acid, b(oh)3 (fm 61.83), should be used to make 2.00 l of 0.050 0 m solution? what kind of fl ask is

grigory [225]

Answer:

Mass = 6.183 g

Solution:

Step 1: Calculate number of moles of Boric acid using following formula,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 0.05 mol.L⁻¹ × 2.0 L

Moles = 0.1 mol

Step 2: Calculate Mass of Boric Acid using following formula,

Moles = Mass ÷ M.mass

Solving for Mass,

Mass = Moles × M.mass

Putting values,

Mass = 0.1 mol × 61.83 g.mol⁻¹

Mass = 6.183 g

Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.

7 0

2 years ago