A solution is made of 12 g NaCl and 70.0 g water. What is the % NaCl?

Explain the importance of concentration as it pertains to a solution

julia-pushkina [17]

There it is, I hope it gets to be helpful

5 0

3 years ago

The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching it

ollegr [7]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

$5 \times 10^{26} atoms = 1 [tex]m^{3}$

2 atoms = $\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms$

= $4 \times 10^{-27} m^{3}$

Formula for volume of a cube is $a^{3}$ . Therefore,

Volume of the cube = $4 \times 10^{-27} m^{3}$

As lattice constant (a) = $(4 \times 10^{-27} m^{3})^{\frac{1}{3}}$

= $1.59 \times 10^{-9} m$

Therefore, the value of lattice constant is $1.59 \times 10^{-9} m$ .

And, for bcc unit cell the value of radius is as follows.

r = $\frac{\sqrt{3}}{4}a$

Hence, effective radius of the atom is calculated as follows.

r = $\frac{\sqrt{3}}{4}a$

= $\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m$

= $6.9 \times 10^{-10} m$

Hence, the value of effective radius of the atom is $6.9 \times 10^{-10} m$ .

3 0

3 years ago

A voltaic cell has a zinc anode and a copper cathode. They are connected by a wire but no salt bridge. What can you predict will

Allisa [31]

The prediction is that B. The electrons will flow to the zinc anode where a negative charge will build up and eventually halt the reaction.

This is known as a chemical element, of the periodic table, that is essential to life and is one of the most widely used metals. Zinc is of considerable commercial importance.

Without the salt bridge, positive and negative charges will build up around the electrodes causing the reaction to stop.

Hence, we know that the purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another.

Read more about<em> zinc</em> here:

brainly.com/question/25764334

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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

What is 11/6 + 11/6 in fractions

Andreyy89

3 2/3 is your answer

6 0

2 years ago

Read 2 more answers