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3 years ago

if 0.683 grams of silver chloride is produced how much (mass) silver nitrate would need to be reacted

1 answer:

4 0

The (mass) silver nitrate would needed to be reacted is calculated as follow

by assuming that AgNo3 reacted with Bacl

that is 2AgNo3 + BaCl2 ---> 2Agcl + Ba(No3)2

find the moles AgNo3 = 0.683g/169.87 g/mol =4.02 x10^-3 moles

by use of mole ratio between Agcl and Agno3 which is 2:2 this implies that the moles of AgNO3 also 4.02 x10^-3

mass= moles x molar mass

= (4.02 x10^-3) x 143.37= 0.576 grams

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3 years ago

A silicon atom has an atomic number of 14. What information does the atomic number tell you? (Choose all possible answers)

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Answer: 90 degrees.

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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.

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3 years ago

Read 2 more answers

What neutral element has the electron configuration described by the Aufbau diagram above.

devlian [24]

Answer:

Aluminium.

Explanation:

The above electronic configuration can be written in a simplified form as shown below:

1s² 2s²2p⁶ 3s²3p¹

Next, we shall determine the number of electrons in the atom of the element as follow:

Number electron = 2 + 2 + 6 + 2 + 1

Number of electron = 13

Next, we shall determine the number of protons.

Since the element is in its neutral state,

The number of electrons and protons are equal i.e

Proton = Electron

Number of electron = 13

Proton = Electron = 13

Proton = 13

Next, we shall determine the atomic number of the element.

The atomic number of an element is simply the number of protons in the atom of the element i.e

Atomic number = proton number

Proton = 13

Atomic number = 13

Comparing the atomic number of the element with those in the periodic table, the element with the above electronic configuration is aluminium since no two elements have the same atomic number.

5 0

3 years ago