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3 years ago

Wich statement best describes the difference between strong nuclear forces and weak nuclear forces

Chemistry

1 answer:

Vika [28.1K]3 years ago

4 0

Answer: option D is right.

Explanation: There are four types of fundamental forces in nature.These are named as gravitational force,electromagnetic force,strong nuclear force and weak nuclear force.

As per the question we have to understand the role of strong nuclear force and weak nuclear force.

An atom consists of a nucleus surrounded by extra nuclear part consisting of electrons in various orbits.The nucleus contains two basic particles called protons and neutrons .Protons are positively charged while neutrons are neutral.Protons being positively charged will impart repulsive force on each other and may come out of the nucleus.But the nucleus is stable.That is due to the strong nuclear force.

Strong nuclear force is a spin dependent and charge independent force which comes into existence due to the mutual interaction of gluons which binds the protons and neutrons .Hence it is attractive in nature.It's 100 times more stronger than electromagnetic force also.

Weak nuclear force comes into existence during radio -active decay .This force is due to the exchange of ' w' and 'z' bosons[the particles like protons and neutrons having integral or zero spin] which are heavier in nature.The role of it is to change protons into neutrons and vice versa.Its a short range force.

Hence the option D is right.

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The equilibrium constant, Kc , for the decomposition of COBr2 COBr2(g) ↔ CO(g) + Br2(g) is 0.190. What is Kc for the following r

charle [14.2K]

Answer:

$2CO(g)+2Br_2(g)\rightleftharpoons 2COBr_2(g)$

27.70 is $K_c'$ for the reaction.

Explanation:

$COBr_2(g)\rightleftharpoons CO(g) + Br_2(g)$

The equilibrium constant of the reaction = $K_c=0.190$

The expression of equilibrium constant is given by :

$K_c=\frac{[CO][Br_2]}{[COBr_2]}$ ..[1]

$2CO(g)+2Br_2(g)\rightleftharpoons 2COBr_2(g)$

The equilibrium constant expression for above reaction can be written as:

$K_c'=\frac{[COBr_2]^2}{[CO]^2[Br]^2}$

$K_c'=\frac{1}{(K_c)^2}$ ( from [1])

$K_c'=\frac{1}{(0.190)^2}=27.70$

27.70 is $K_c'$ for the reaction.

7 0

4 years ago

Lakes can disappear due to _

cupoosta [38]

Answer:

i think it's eutrophication

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3 years ago

Explain percent ionic character of a bond. Explain percent ionic character of a bond. Percent ionic character is defined as the

disa [49]

Answer:

Percent ionic character is defined as the ratio of a bond’s actual dipole moment to the dipole moment it would have if the electron were completely transferred from one atom to the other, multiplied by 100

Explanation:

The percent ionic character seeks to establish the amount of electrovalency in a particular compound. It simply compares the covalency i.e extent of shared electrons to the the electrons that are transferred.

It is given as the ratio of the acutal dipole moment to the dipole moment due to ionic character of the bond multiplied by 100:

Percent ionic character = μ_obs/μ_ionic x 100

Where μobs is the actual dipole moment and μ ionic is the dipole moment if the bonds were 100% ionic.

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3 years ago

A flashlight has a resistance of 2.4 . What voltage is applied by the batteries if the current in the circuit is

hammer [34]

Answer : The voltage applied by the batteries is, 6.0 V

Solution : Given,

Resistance of flashlight = 2.4 ohm

Current in the circuit = 2.5 Ampere

Formula used :

$V=IR$

where,

V = applied voltage

I = current in the circuit

R = resistance of light

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

$V=(2.5A)\times (2.4ohm)=6volt=6V$

Therefore, the voltage applied by the batteries is, 6.0 V

6 0

3 years ago

Read 2 more answers

A lab analysis of 0.502 g of Scandium (Sc) reaction with Oxygen (O2) from Air resulted

VARVARA [1.3K]

The empirical formula of the metal oxide would be $Sc_2O_3$

<h3>What is empirical formula?</h3>

Mass of scandium = 0.502g

Mass of scandium + oxygen = 0.768g

Mass of oxygen = 0.768 - 0.502 = 0.266g

Mole of scandium = 0.502/44.95 = 0.011

Mole of oxygen = 0.266/16 = 0.017

Divide by the lowest:

Scandium = 0.011/0.011 = 1

Oxygen = 0.017.0.011 = 1.5

Thus, the empirical formula of the compound would be $ScO_1.5$ or $Sc_2O_3$

More on empirical formula can be found here:brainly.com/question/14044066

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3 years ago