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Kisachek [45]
3 years ago
5

Help ASAP ......begdhzhshs

Chemistry
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

thunderstorms

Explanation:

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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
Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 0.855g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal = 39.09 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell = ?

Putting values in above equation, we get:

0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm

<u>Conversion factor:</u>  1cm=10^{10}pm

Hence, the edge length of the crystal is 533.5 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm

Hence, the atomic radius of potassium is 231.01 pm

4 0
3 years ago
Which intermolecular force is common to all polar molecules but not nonpolar molecules? dispersion forces dipole-dipole forces h
Schach [20]
Answer:
            <span>Dipole-Dipole Forces are common </span><span>to all polar molecules but not non polar molecules.

Explanation:
                   An Asymmetrical molecule having a region of high electron density (partial negative) and lower density (partial positive) interacts with its neighbor molecules through Dipole-Dipole Interactions. The partial positive part of one molecule interacts with the partial negative part of another.

Example:
               Acetone having a partial positive carbon (of carbonyl group) and partial negative oxygen interacts through Dipole-Dipole forces. Hence, acetone does not involves Hydrogen Bond interactions.</span>
5 0
4 years ago
What is rusting? Write its chemical equations.​
anyanavicka [17]

\huge \quad\quad \underline{ \tt{{Answer}}}

Rusting of iron is the most familiar example of corrosion. It is a process in which an iron reacts with atmospheric oxygen and moisture to form a reddish brown substance iron oxide, which is commonly known as rust.

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

\blue{ \star}:\sf4fe+ 3O_2+ XH_2O→ 2Fe_2O_3.XH_2O

5 0
2 years ago
Read 2 more answers
Copper and silver are examples that will form what bond
Helen [10]

A metallic bond since both are pure metals but are not ionic



6 0
4 years ago
In order to complete his research project, Roger needs to make a mixture of 86 mL of a 36% acid solution from a 39% acid
Zina [86]

The volume of the 39% acid solution that Roger needs to use to make a mixture of 86 mL of a 36% acid solution is 74.27 mL.

The mixture of the acids solutions is given by:

CV = C_{1}V_{1} + C_{2}V_{2}    (1)      

Where:

C: is the concentration if the mixture = 36%  

V: is the total volume of the mixture = 86 mL

C₁: is the concentration of acid 1 = 39%

V₁: is the volume if acid 1 =?

C₂: is the concentration of acid 2 = 17%

V₂: is the volume of acid 2

The sum of V₁ and V₂ must be equal to V, so:

V = V_{1} + V_{2}

V_{2} = V - V_{1}  (2)

By entering equation (2) into (1), we have:

CV = C_{1}V_{1} + C_{2}(V - V_{1})

36\%*86 mL = 39\%*V_{1} + 17\%(86 mL - V_{1})

Changing the percent values to decimal ones:

0.36*86 mL = 0.39*V_{1} + 0.17(86 mL - V_{1})  

Now, by solving the above equation for V₁:

V_{1} = 74.27 mL  

Therefore, the volume of the 39% acid solution is 74.27 mL.

       

To learn more about mixture and solutions, go here: brainly.com/question/6358654?referrer=searchResults

I hope it helps you!  

         

7 0
3 years ago
Read 2 more answers
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