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3 years ago

The energy released by a nuclear reaction results primarily from the

2 answers:

8 0

Conversion of mass into energy. When an atom is split, the mass will be slightly less, and this weight difference will be converted into energy. I hope that helps!

Anni [7]3 years ago

7 0

Answer:

Conversion of mass into energy

Explanation:

In chemical reactions, the mass is conserved. In chemical reaction the matter cannot be created nor be destroyed.

While in nuclear reaction, the mass is not conserved. The mass gets converted to energy.

The relation is : E= mc²

Thus in nuclear reactions tremendous amount of energy is released.

In both fission and fusion reactions mass gets converted to energy.

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You come home and find that there's been a power outage while food was cooking in the slow cooker. What should you do wit the fo

klasskru [66]

Answer:

Well you can not serve it because it is a slow cooker it cooks slow and the refrigerator uses power.Stove tops use power too, I wouldn't feed it to the dog because it whatever you made could possibly make the dog sick and throwing it away is a waste of money. Out of all of those explanations choose the best one. Which ever one you pick whether right or wrong, think about it.

Explanation:

7 0

3 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

Complete and balance the following neutralization reaction Ca(OH)2 + H2SO4

Debora [2.8K]

Answer: Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + 2H2O(l)

Explanation:

Since this is a neutralization reaction, the end product would be salt and water. In this equation Calcium will displace hydrogen from the acid because it is more reactive, resulting in the formation of CaSO4 (salt), while the displaced H2 molecule combines with OH molecules to form water.

The equation of the reaction is thus;

Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + H2O(l), in other to balance it, we add ''2'' to the water molecule in the right hand side of the equation.

Balance equation is

Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + 2H2O(l)

6 0

2 years ago

Explore: Arrange the matter

Trava [24]

Answer:

The levels, from smallest to largest, are: molecule, cell, tissue, organ, organ system, organism, population, community, ecosystem, biosphere.

Explanation:

The biological levels of organization of living things arranged from the simplest to most complex are: organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystem, and biosphere.

7 0

2 years ago

Read 2 more answers

(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t

11Alexandr11 [23.1K]

Answer:

The radius of an impurity atom occupying FCC octahedral site is $0.414{\rm{R}}$

The radius of an impurity atom occupying FCC tetrahedral site is $0.225{\rm{R}}$ .

Explanation:

In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals

An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.

Step 1 of 2

(1)

The position of an atom that fits in the octahedral site with radius $\left( r \right)$ is as shown in the first uploaded image.

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The radius of the impurity is as follows:

$2r=a-2R------(A)$

The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:

Consider $\Delta {\rm{XYZ}}$ as follows:

$(XY)^ 2 =(YZ) ^2 +(XZ)^2$

Substitute $XY$ as ${\rm{R}} + 2{\rm{R + R}}$ and ${\rm{YZ}}$ as a and ${\rm{ZX}}$ as a in above equation as follows:

$(R+2R+R) ^2 =a ^2 +a^ 2\\16R ^2 =2a^ 2\\ a =2\sqrt{2R}$

Substitute value of aa in equation (A) as follows:

$r= \frac{2\sqrt{2}R -2R }{2} \\ =\sqrt{2} -1R\\ = 0.414R$

The radius of an impurity atom occupying FCC octahedral site is $0.414{\rm{R}}$

Note

An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.

Step 2 of 2

(2)

The impure atom in FCC tetrahedral site is present at the body diagonal.

The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The body diagonal is represented by AD.

The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:

$AD=2R+2r----(B)$

In $\Delta {\rm{ABC}},$

$(AB) ^2 =(AC) ^2 +(BC) ^2$

For calculation of AD, AB is determined using Pythagoras theorem.

Substitute ${\rm{AC}}$ as a and ${\rm{BC}}$ as a in above equation as follows:

$(AB) ^2 =a ^2 +a ^2$

$AB= \sqrt{2a} ----(1)$

Also,

$AB=2R$

Substitute value of $2{\rm{R}}$ for ${\rm{AB}}$ in equation (1) as follows:

$2R= \sqrt{2} aa = \sqrt{2} R$

Therefore, the length of body diagonal is calculated using Pythagoras Theorem in $\Delta {\rm{ABD}}$ as follows:

$(AD) ^2 =(AB) ^2 +(BD)^2$

Substitute ${\rm{AB}}$ as $\sqrt 2$ a and ${\rm{BD}}$ as a in above equation as follows:

$(AD) ^2 =( \sqrt 2a) ^2 +(a) ^2 AD= \sqrt3a$

For calculation of radius of an impure atom in FCC tetrahedral site,

Substitute value of AD in equation (B) as follows:

$\sqrt 3a=2R+2r$

Substitute a as $\sqrt 2$ ${\rm{R}}$ in above equation as follows:

$( \sqrt3 )( \sqrt2 )R=2R+2r\\\\$

$r = \frac{2.4494R-2R}{2}\\$

$=0.2247R$

$\approx 0.225R$

The radius of an impurity atom occupying FCC tetrahedral site is $0.225{\rm{R}}$ .

Note

An impure atom occupies the tetrahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The length of body diagonal is calculated using Pythagoras Theorem. The body diagonal is equal to the sum of the radii of two atoms. This helps in determining the relation between the radius of impure atom and radius of atom present in the unit cell.

7 0

3 years ago