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3 years ago

Explain and describe how the photoelectric effect occurs on an atomic level in terms of protons, neutrons, and electrons

Chemistry

1 answer:

grin007 [14]3 years ago

3 0

Answer:

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons.

Based on the wave model of light, physicists predicted that increasing light amplitude would increase the kinetic energy of emitted photoelectrons, while increasing the frequency would increase measured current.

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Given 1kg=2.20 lbs, how many kilograms are in 125 lbs? (Metric conversion)

Nadya [2.5K]

Answer:

56.82 Kg

Explanation:

Given data:

1 Kg = 2.20 lbs

Number of kilogram in 125 lbs = ?

Solution:

lbs is used for pound. lb is abbreviation of libra. It is Latin word meaning balance.

Both kilogram and pounds are units of mass. Pound is smaller unit than kilogram.

one Kg = 2.20 lbs

Number of kg in 125 lbs:

125 lbs × 1 Kg/2.20 lbs

125 lbs.Kg/2.20 lbs

56.82 Kg

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3 years ago

What does it mean to truly love someone?

Minchanka [31]

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2 years ago

A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

Semenov [28]

Answer: The correct answer is Option d.

Explanation:

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$