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morpeh [17]
3 years ago
5

Calculate the nuclear binding energy in mega-electronvolts (MeV) per nucleon for Th232. Th232 has a nuclear mass of 232.038 amu.

Chemistry
1 answer:
SOVA2 [1]3 years ago
3 0

Answer:

1720 MeV

Explanation:

Thorium has 90 protons and 142 neutrons

The mass of each proton is 1.007277 amu

The mass of each neutron is 1.008665 amu

The mass of 90 protons is 90 * 1.007277 amu = 90.65493 amu

The mass of 142 neutrons is 142 * 1.008665 amu = 143.23043 amu

Total calculated mass of nucleons = 90.65493 amu + 143.23043 amu = 233.88536

Mass defect = 233.88536 amu - 232.038 amu = 1.84736 amu

We convert from amu to MeV using the conversion factor of 931

Binding energy = 1.84736 amu * 931 = 1720 MeV

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Explanation:

The coiling up of DNA with the help of histone proteins - to what is called heterochromatic regions-  is part of gene regulation. This makes genes inaccessible to RNA polymerase that makes mRNA from the genes. When the genes are exposed by DNA unwinding, these genes are transcribed and the resulting mRNAs are translated by ribosomes into proteins.

The DNA never unwinds completely, but rather does so region by region, because if it does so it would become so long that it wouldn't fit in the nucleus or cell.

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3 years ago
1.The movement of particles from a less crowded area to a more crowded area requires
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4 years ago
Rank the following hydroxides in order of increasing aqueous basicity: Al(OH)₃, (OH)₃, In(OH)₃
vovangra [49]

The correct answer is B(OH)3 <AI(OH)3 < In(OH)3.

The metallic character of the group 1 elements first increases from Boron to Aluminium, then decreases from Aluminium to Thallium because of high ionization Enthalpy. Also, the larger size of the ion, the lesser is the ionization of Enthalpy.

Basic nature of the hydroxides of group 13 increases on moving down the group as the electro-positive character of elements increases.

Therefore, the correct order of increasing aqueous basicity is as follows:

B(OH)3 <AI(OH)3 < In(OH)3

What is the basic nature of group 13?

The metallic charecter of the elements affects their fundamental characteristics. The basicity rises as the metallic character does, and vice versa. Therefore, as we move lower in any given group, the atomic radius of the elements increases. Higher radii indicate that it would become increasingly challenging for the nucleus to rule over the electrons in the valency shell. Consequently, there will be a greater inclination for electron release. As a result, the metallic properties will improve, increasing the basicity of the oxides. Therefore, it follows that the basicity of the oxides would rise as we go down in a group.

To learn more about hydroxides refer the link:

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7 0
2 years ago
Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to
Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density \rho_{avg}; we have:

\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }

The average atomic weight is:

A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

C_{Fe} = 100 - C_v

From a unit cell volume V_c

V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}

where;

N_A = number of Avogadro constant.

SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

C_{Fe} with 100-C_v

Then:

a^3 = \dfrac   { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) }    {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

Replacing the values; we have:

(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol)   }   }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3)   } }

2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} }  \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }

2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})

\mathbf{C_v = 9.1 \ wt\%}

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