All categories

3 years ago

Of the atoms below, __________ is the most electronegative. PLZ HELP I'LL AWARD BRINLIEST.

Chemistry

1 answer:

Readme [11.4K]3 years ago

8 0

Hello sir, The answer that you are looking for is fluorine. Fluorine is the most electronegative atom. I am also sorry for calling you sir. I didn't know you were a woman.

You might be interested in

Isoflavones, quercetin and anthocyanins are all what type of phytochemicals

kykrilka [37]

Isoflavones, quercetin, and anthocyanin are all Flavonoids.

8 0

3 years ago

The factor that is changed throughout an experiment is called the _______. A. apparatus B. constant C. variable D. hypothesis

ser-zykov [4K]

The variable is what changes during an experiment. Hopefully this helped! :)

4 0

3 years ago

Read 2 more answers

Which particle in an atom is lost or gained to make an ion

denpristay [2]

An electron because that is the only part able to be lost or gained without nuclear action needed

7 0

3 years ago

using the Bohr model for hydrogen: energy = hc/wavelength = 2.18 x 10^-18 Joules (1/nf2 - 1/ni2) N=15 to n=5

soldier1979 [14.2K]

Answer:

Energy lost is 7.63×10⁻²⁰J

Explanation:

Hello,

I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5

E = hc/λ(1/n₂² - 1/n₁²)

n₁ = 15

n₂ = 5

hc/λ = 2.18×10⁻¹⁸J (according to the data)

E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)

E = 2.18×10⁻¹⁸ (1/15² - 1/5²)

E = 2.18×10⁻¹⁸ ×(-0.035)

E = -7.63×10⁻²⁰J

The energy lost is 7.63×10⁻²⁰J

Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level

5 0

3 years ago

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

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

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago