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Iteru [2.4K]
3 years ago
10

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

(a) the lattice parameter; and (b) the atomic radius of potassium
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
7 0

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8 cm

for a BCC structure, the atomic radius is equal to

r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3}  }{4}=2.3103x10^{-8}cm

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iris [78.8K]
The answer is D, wavelength

Hope I helped!
7 0
3 years ago
The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?
Vedmedyk [2.9K]

Answer : The volume of pure diamond is 0.493inch^3

Explanation : Given,

Density of pure carbon in diamond = 3.52g/cm^3

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}

Molar mass of carbon = 12 g/mol

\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

Density=\frac{Mass}{Volume}

Now putting all the given values in this formula, we get:

3.52g/cm^3=\frac{284.4g}{Volume}

Volume = 80.8cm^3

As we know that:

1cm^3=0.061inch^3

So,

Volume = 0.061\times 80.8inch^3

Volume = 0.493inch^3

Therefore, the volume of pure diamond is 0.493inch^3

5 0
3 years ago
A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c
pshichka [43]
<span>Answer: Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025 Ca(NO3)2 >> Ca2+ + 2NO3- Moles NO3- = 2 x 0.025 = 0.05 Moles HNO3 = 400 x 0.100 / 1000 = 0.04 Total moles = 0.05 + 0.04 = 0.09 Total volume = 500 ml = 0.500 L M = 0.09 / 0.500 = 0.18</span>
6 0
3 years ago
A student used 1.168g of an unknown weak acid and titrated it against 28.75 mL of 0.105M NaOH to reach the equivalence point. Wh
Vaselesa [24]

Answer:

The molar mass of the unknown acid is 386.8 g/mol

Explanation:

Step 1: Data given

Mass of the weak acid = 1.168 grams

volume of NaOH = 28.75 mL = 0.02875 L

Molarity of NaOH = 0.105 M

Since we only know 1 equivalence point, we suppose the acid is monoprotic

Step 2: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.105 M * 0.02875 L

Moles NaOH = 0.00302 moles

We need 0.00302 moles of weak acid to neutralize the NaOH

Step 3: Calculate molar mass of weak acid

Molar mass = mass / moles

Molar mass = 1.168 grams / 0.00302 moles

Molar mass = 386.8 g/mol

The molar mass of the unknown acid is 386.8 g/mol

3 0
3 years ago
Be(OH)2 is solid why?​
pickupchik [31]

Answer:

no it's not solid rather it's an aqueous

Explanation:

B/c Barium hydroxide is used in analytical chemistry for the titration of weak acids, particularly organic acids. Its clear aqueous solution is guaranteed to be free of carbonate, unlike those of sodium hydroxide and potassium hydroxide, as barium carbonate is insoluble in water.

8 0
2 years ago
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