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1 year ago

Light of frequency 3.62 x 1015 Hz

Chemistry

1 answer:

vladimir1956 [14]1 year ago

8 0

The kinetic energy that the liberated electron has in eV based on the given values is :

10.28 eV.

<h3>Explanation:</h3>

1 eV is the kinetic energy gained by an electron or proton when acted upon by a 1 volt potential difference. E = QV is the formula for energy in terms of charge and potential difference.

A photoelectron's maximum kinetic energy is given by

Emax = hf - W

where h is Plank's constant,

f is the incident photon's frequency, and

W denotes the metal surface's work function.

here given,

f = 3.62 x 10¹⁵ Hz

W = 4.70 eV

by recalling the equation for a photoelectron's maximum kinetic energy with given values,

Emax = hf - W

= ((4.14 × 10⁻¹⁵eV.s)(3.62 x 10¹⁵ Hz)) - (4.70 eV)

= 10.28 eV

As a result, we discovered that the maximum kinetic energy of electrons is 10.28eV.

To learn more about kinetic energy of electrons refer to :

brainly.com/question/21208918

#SPJ13

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2 years ago

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This would be 1.22 x 10^1

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2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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3 years ago

Describe the relationship between the radius of a cation and that of the atom from which it is formed. a. cations are much small

koban [17]

<h3><u>Answer;</u></h3>

Cations are much smaller than their corresponding parent

<h3><u>Explanation;</u></h3>

Parent atom has more electrons and thus the effective nuclear charge on each electron is less.
When a cation is formed electron(s) is/are lost. Thus the effective nuclear charge or simply put, the attraction of the nucleus towards the electrons increases. Therefore, due to greater pull, the nucleus pulls the shells towards it, there by reducing the size, which makes cations smaller than their corresponding parent.

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2 years ago

Select all the correct images.<br> Select the atomic models that belong to the same element.

Alex_Xolod [135]

Answer:The 2nd and 3rd one.

Explanation:

It has the same number of protons but different amount of nuetrons.

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2 years ago