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3 years ago

Help please! I don’t understand this

Chemistry

1 answer:

AlexFokin [52]3 years ago

5 0

I don’t really know how to explain how to write electron configurations in a short paragraph so maybe look up a video? It’s really easy to understand once you get the hang of it but i’ll attach a picture to help you write electron configurations for any element. Try to think of it as reading the periodic table from left to write until you end up on the element.

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If the atomic weight of nitrogen is 14.01, what is the mass of the nitrogen atoms in one mole of cadmium nitrate, Cd(NOÀ)

Nadya [2.5K]

The answer is 28.02.

The chemical formula for cadmium nitrate is Cd(NO₃)₂.

There are in total 2 nitrogen atoms in Cd(NO₃)₂. If the atomic weight of nitrogen is 14.01, the mass of two nitrogen atoms in one mole of cadmium nitrate is 28.02:
2 · 14.01 = 28.02

8 0

3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

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

where,

$\rho$ = density = $10.28g/cm^3$

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

M = atomic mass of metal (molybdenum) = 95.94 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:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

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 = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

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3 years ago

Compared to micellular Compound 1, Compound 2 is structurally more rigid as a result of what type of interaction?

zimovet [89]

Answer:

D. Intramolecular covalent bond

Explanation:

Compound D is structurally more rigid as a result of intramolecular covalent bonding. The forces that hold together atoms within a compound are greater as compared to forces holding two molecules together (intermolecular bonding). On the other hand Hydrogen bonds are weaker as compared to covalent bonds. Covalent bonds involve the sharing of electrons between two atoms and Hydrogen bonds are formed between a highly electronegative atom like oxygen, Flourine,Chlorine to hydrogen.

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3 years ago

There are two opposing processes that occur in a solution in contact with undissolved solute. These are dissolving and ________.

Alex777 [14]

Answer:

c. crystallization

Explanation:

The opposing process that occur in a solution in contact with undissolved solute are dissolution and crystallization.

In the dissolution process the solid substance coverts into liquid state and mixes with solution. Whereas in Crystallization the the chemical is converted from the liquid solution to solid crystal state.

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3 years ago

Some people say that only gay and bisexual men are likely to be infected with hub true false

shusha [124]

Answer:

They do have a larger chance, but others might be infected too.

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