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3 years ago

Which statement is true at STP? (The atomic mass of Zn is 65.39 u.)

1 answer:

7 0

Zinc is a metal. At STP, it exists as solid and is stable as it is. It is an important mineral and is used in many applications like in food, metal and drugs. Zinc can be found in the Earth's crust and also it is present in small amounts in some food.

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Calculate the ratio of naf to hf required to create a buffer with ph = 4.05.

NARA [144]

PH is the test of acidity or basicity of a solution. it follows the formula:
pH = pKa + log [salt] / [acid] where NaF is the salt and HF is the acid in this case.

By literature, Ka of HF is 3.5*10^-4
pKa= -log(Ka)= 3.46 

pH = pKa + log [NaF / [HF] 

4.05 = 3.46 + log [NaF / [HF]

log [NaF / [HF] = 0.59

[NaF / [HF] = 3.89

4 0

3 years ago

Read 2 more answers

8.43g of CaCl2·2H2O is dissolved in 40.0g of H2O. What is the molality of the solution? (Note that the H2O in the CaCl2·2H2O mus

lara31 [8.8K]

Answer:

[CaCl₂·2H₂O] = 1.43 m

Explanation:

Molality is mol of solute / kg of solvent.

Mass of solvent = 40 g

Let's convert g to kg → 40 g / 1000 = 0.04 kg

Let's determine the moles of solute (mass / molar mass)

8.43 g / 146.98 g/mol = 0.057 mol

Molality = 0.057 mol / 0.04 kg → 1.43

3 0

3 years ago

Read 2 more answers

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,

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