The ability of an atom to attract the shared electrons in a covalent bond is its

The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

$v^2=0.869446\times 10^{12}$

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

5 0

3 years ago

The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.

exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. $Na_{2} CO_{3}$

Major Species at B:

1. $Na_{2} CO_{3}$

2. $NaHCO_{3}$

Major Species at C:

1. $NaHCO_{3}$

Major Species at D:

1. $NaHCO_{3}$

2. $H_{2}CO_{3}$

Major Species at E:

1. $H_{2}CO_{3}$

Major Species at F:

1. $H_{2}CO_{3}$

b) pH calculation:

At Halfway point B:

pH = pK $a_{1}$ + log[ $CO_{3}$ $.^{-2}$ ]/[H $CO_{3}$ $.^{-1}$ ]

pH = pK $a_{1}$ = 6.35

Similarly, at halfway point D.

At point D,

pH = pK $a_{2}$ + log [H $CO_{3}$ $.^{-1}$ ]/[H2 $CO_{3}$ ]

pH = pK $a_{2}$ = 10.33

8 0

3 years ago

Control activities are performed within processes (e.g., segregation of duties required in processing cash receipt transactions)

galina1969 [7]

Answer:

a. True

Explanation:

The control objectives seek to support the fulfillment of the critical factors of business success, in order to finally support the achievement of the institutional objectives, on this the CITI is based

It is very important to clearly identify the relationships of the control objectives with other internal control elements such as:

· Critical success factors receive direct support from control objectives to support compliance.

· The control objectives are made up of control goals, which allow them to have a more detailed perspective of the control objectives and at the same time facilitate the evaluation of their compliance.

CITI CONCEPT:

It is the set of administration elements that a company establishes in a coordinated way so that the use of information technology resources effectively supports the institutional objectives of the company

6 0

3 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0

3 years ago

Unstable isotopes undergo radioactive decay. what occurs during radioactive decay?

mel-nik [20]

Here is what radioactive decay is:
<span>Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.</span>

5 0

3 years ago