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3 years ago

How does the mass-energy equation E = mc2 relate to fission?

Chemistry

2 answers:

balu736 [363]3 years ago

7 0

It relates because an object requires fission and energy to its mass which is why energy equals mass times csquared

kodGreya [7K]3 years ago

3 0

Answer:

Check explanation

Explanation:

First, let us take a look at the meaning of Nuclear fission, Mass and Energy.

MEANING OF NUCLEAR FISSION: Nuclear fission can be defined as the breaking down of large atom into smaller with the release of energy.

MEANING OF MASS: Mass is a physical property which shows the resistance to acceleration when force is applied.

MEANING OF ENERGY: energy is a property that must transferred from one body to another.

E= MC^2 ---------------------------------(1).

Where E= energy released, M= mass, and C= speed.

The equation (1) above is the Albert Einstein's equation.

During fission when large atoms split into small ones. Here, we have a decrease in the mass of the atomic nucleus. This means that mass has been converted into energy.

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The concentrations of Fe and K in a sample of riverwater are 0.0400 mg/kg and 1.30 mg/kg, respectively. Express the concentratio

Ierofanga [76]

Answer :

The concentration in molality of Fe is, $7.1\times 10^{-7}mol/kg$

The concentration in molality of K is, $3.3\times 10^{-5}mol/kg$

Explanation:

First we have to calculate concentration in molality of Fe.

Molar mass of Fe = 56 g/mol

Concentration of Fe = 0.0400 mg/kg = $4\times 10^{-5}g/kg$

Conversion used : 1 g = 1000 mg

$\text{Concentration in molality}=7.1\times 10^{-7}mol/kg$

Thus, the concentration in molality of Fe is, $7.1\times 10^{-7}mol/kg$

Now we have to calculate concentration in molality of K.

$\text{Concentration in molality}=\frac{\text{Concentration of K}}{\text{Molar mass of K}}$

Molar mass of K = 39 g/mol

Concentration of K = 1.30 mg/kg = $1.3\times 10^{-3}g/kg$

Conversion used : 1 g = 1000 mg

$\text{Concentration in molality}=\frac{1.3\times 10^{-3}g/kg}{39g/mol}$

$\text{Concentration in molality}=7.1\times 10^{-7}mol/kg$

Thus, the concentration in molality of K is, $3.3\times 10^{-5}mol/kg$

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3 years ago

What are observable changes when burning a log?

Irina-Kira [14]

Log being dissolved and sparkes flaring out

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4 years ago

Read 2 more answers

Click oil basics to explore the natural resources needed for the production of plastics. use the information to answer the quest

GaryK [48]

The ingredients that is important in making plastics is natural gas plant liquids.

Natural gas liquids (NGLs) are components of natural gas which are separated from the gas state in the form of liquids. They many applications including; plastic production, inputs for petrochemical plants, they are also burned for space heat and cooking, and can also be blended into vehicle fuel.

In the manufacturing of plastics, come components of NGLs are used. And for this case, ethane. Ethane is used in the production of ethylene which are passed through pressure and catalyst to be turned to plastics like polythene.

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Learn more about natural gas liquids here:

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8 0

2 years ago

Sebanyak 4,5 mg magnesium hidroksida (Mg(OH)2) dapat larut dalam 500 ml air. nyatakan kelarutan Mg(OH)2 dalam mol L-1

Schach [20]

S =gr/mr x 1000/v s=4,5/58 x 1000/500 s=4,5/58 x 2 s=9/58 s= 0,1

8 0

4 years ago

The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

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4 years ago