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3 years ago

What is the main purpose of the second paragraph?

Chemistry

1 answer:

Svetlanka [38]3 years ago

5 0

To explain your first paragraph which includes your thesis. The second paragrph supports your first pragraph

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

3 years ago

The molecular mass of the compound is 132 amu. What is the molecular formula?

densk [106]

C6H12O3 is a molecular formula that contains 54.5% C, 9.1% H, and 36.4% O and <span>has a molar mass 132 amu. </span>

4 0

3 years ago

How many liters of C3H6O are present in a sample weighing 25.6 grams?

lawyer [7]

To Find :

Number of moles of C₃H₆O present in a sample weighing 25.6 grams.

Solution :

Molecular mass of C₃H₆O is :

M = (6×12) + (6×1) + (16×1) grams

M = 94 grams/mol

We know, number of moles of 25.6 grams of C₃H₆O is :

$n = \dfrac{Given \ Mass \ Of \ C_3H_6O }{Molar\ Mass \ Of \ C_3H_6O }\\\\n = \dfrac{25.6}{94}\ mole\\\\n = 0.27 \ mole$

Hence, this is the required solution.

4 0

3 years ago

Which of the following is a scientific question? O A. How can I make a peach cobbler? O B. What chemicals cause most plants to b

igor_vitrenko [27]

The answer is B. Because think about it, it’s not a scientific fact that brown dogs are better pets, and the best color for your room isn’t it because the answer would be an opinion of someone not a fact. I hope this might help you.

5 0

3 years ago

One method of removing phosphate from wastewater effluents is to precipitate it with aluminum sulfate. A plausible stoichiometry

Viktor [21]

Answer: The equation is written below.

Explanation:

$2PO_4^{3-}(aq)+Al_2(SO_4)_3(aq)\rightarrow 2AlPO_4(s)+3SO_4^{2-}(aq)$

According to Stoichiometry of the reaction:

2 moles of phosphate ions reacts with 1 mole of aluminum sulfate to produce 2 moles of aluminum phosphate precipitate and 3 moles of sulfate ions.

Aluminum phosphate is an odorless and white crystalline solid

The chemical equation is written above.

8 0

3 years ago