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3 years ago

PLEASE HELP BRAINLIEST RN

Chemistry

2 answers:

Mademuasel [1]3 years ago

7 0

Answer: C is. sssss

Explanation:

Gala2k [10]3 years ago

3 0

C is the correct answer

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2. Find the two generic molecules from Part 1 that are made of 3 atoms. a. Compare and contrast these two molecules by listing t

Stells [14]

Answer:

hello the molecules are missing from your question below are the Generic molecules : $ABE_{3}$ and $AB_{3} E$

answer : It can be determined that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

They have different Geometry

They differ in bond angles as well

Explanation:

The two generic molecules : $ABE_{3}$ and $AB_{3} E$

comparing(similarities) these two generic molecules

It can be determined that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

differences between the generic molecules

They have different Geometry

They differ in bond angles as well

3 0

3 years ago

An imaginary element with BCC structure and has an atomic radius of 0.17 nm, with a molar mass of 56.08 g/mol. What is the densi

Oksi-84 [34.3K]

<u>Answer:</u> The density of the given element is $3.07g/cm^3$

<u>Explanation:</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 = 0.17 nm

a = edge length = ?

Putting values in above equation, we get:

$0.17=\frac{\sqrt{3}\times a}{4}\\\\a=\frac{0.17\times 4}{\sqrt{3}}=0.393nm$

To calculate the density of metal, we use the equation:

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

where,

$\rho$ = density

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

M = atomic mass of metal = 56.08 g/mol

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

a = edge length of unit cell = $0.393nm=3.93\times 10^{-8}cm$ (Conversion factor: $1cm=10^{7}nm$ )

Putting values in above equation, we get:

$\rho=\frac{2\times 56.08}{6.022\times 10^{23}\times (3.93\times 10^{-8})^3}\\\\\rho=3.07g/cm^3$

Hence, the density of the given element is $3.07g/cm^3$

3 0

3 years ago

SOMEONE JUST PLEASE HELP MEE

aev [14]

Answer:

A is the answer. Hope this helped.

8 0

4 years ago

Read 2 more answers

At room temperature (208C) and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Ugo [173]

Total density of filled ball with nitrogen gas: $\frac{0.12G+0.6524g}{0.560L} = 1.3792g/L$

The relationship between mass and volume can be easily determined using density; for example, the mass of a body is equal to its volume multiplied by the density (M = Vd), whereas the volume is equal to the mass divided by the density (V = M/d). The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air. Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165g/L$