Please help. Will give brainliest

Leona [35]

The answer is B

To write the equilibrium constant for an equation, all you have to do is divide the products by the reactants. The reactants are always on the left side, and the products are always on the right side. The coefficients of the elements will be written as the exponent of that same element. However, in this equation, we do not have to write any exponents, as there are no coefficient but 1.

5 0

3 years ago

Answer these questions based on 234. 04360 as the atomic mass of thorium-234. The masses for the subatomic particles are given.

nikdorinn [45]

The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.

<h3>How do we calculate atomic mass?</h3>

Atomic mass (A) of any atom will be calculated as:

A = mass of protons + mass of neutrons

In the Thorium-234:

Number of protons = 90

Number of neutrons = 144

Mass of one proton = 1.00728 amu

Mass of one neutron = 1.00866 amu

Mass of thorium-234 = 90(1.00728) + 144(1.00866)

Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu

Given mass of thorium-234 = 234.04360 amu

Mass defect = 235.90224 - 234.04360 = 1.85864 amu

Hence required value is 1.85864 amu.

To know more about Atomic mass (A), visit the below link:

brainly.com/question/801533

4 0

2 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

3 years ago

Read 2 more answers

B) Which term is defined as a measure of the average kinetic energy of the particles in a sample of matter?

kogti [31]

Entropy is also defined as a measure of the average kinetic energy of particles in a sample of matter.

6 0

3 years ago

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What is one of the primary unanswered questions about the origin of the universe? What type of elements formed first? Approximat

torisob [31]

Answer: last option, what came before the big bang?

Explanation:

The big bang theory states that the universe started as a dense nucleus of matter: a huge amount of matter concentrated in a tiny spot.

This is the conclusion of equations and evidences that prove that the universe has been and continuous to expand: since it has been expanding, there was a moment when it was as small and dense as it is possible.

So, the expansion is the result of violent explosion.

The time during which the expansion has been happening (this is how long ago the big bang occured) has been estimated thanks the the observation of the speed of recesion of the galaxies, but nothing can be told about what came before the bing bang occured.

7 0

3 years ago

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