Give the half-equation for the reaction at the anode, during

Electrons are ejected from a surface with speeds ranging up to 1380 km/s when light with a wavelength of 173 nm is used. What is

mixer [17]

Answer:

ΔE(work)= 692 Kj/mole e⁻ = 4.32 x 10²⁴ eV/mole e⁻

Explanation:

ΔE = hc/λ = (6.63 x 10⁻³⁴j·s)(2.99792 x 10⁸m/s)/(1.73 x 10⁻⁷m) = 1.15 x 10⁻¹⁸j/e⁻ x 6.023 x 10²³ e⁻/mole = 691,999 j/mole e⁻ = 692 Kj/mole e⁻ x 6.24 x 10²¹ eV/KJ = 4.32 x 10²⁴ eV/mole e⁻

8 0

3 years ago

The two major classes of telescopes are the _____ and the _____ telescope.

sukhopar [10]

Answer:

refractors and reflectors

Explanation:

4 0

3 years ago

Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)

vichka [17]

Answer:

Kb → 1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m = Kb → 1.56 °C / m

3 0

3 years ago

What are the formal charges of both chlorines and iodine in icl2?

sergejj [24]

Answer:

Explanation:

Formal charge of ICl₂⁻

Formal charge = group no - ( no of non bonding electrons +no of bonds)

In I there are 7 electrons in outermost orbit . If we add one more electrons due to - ve charge on the ion , it becomes eight . This centrally placed iodine forms two single bond with two chlorine atoms on either side.

Each of chlorine atoms also contains 7 valance electrons like iodine.

So formal charge of chlorine

= group no - ( no of non bonding electrons +no of bonds)

= 7 - ( 6 + 1 )

= 0

So formal charge of iodine

= group no - ( no of non bonding electrons +no of bonds)

= 7 - ( 5 + 2 )

=0

Formal charge of ICl₂⁺

In this case , central iodine will have only 6 valence electrons due to absence one electron.

So formal charge of chlorine in ICl₂⁺

= group no - ( no of non bonding electrons +no of bonds)

= 7 - ( 6 + 1 )

= 0

formal charge of iodine in in ICl₂⁺

7 - ( 4 + 2)

= 1

6 0

4 years ago

SOMEONE PLEASE HELP ME WITH THIS QUESTION!!!!!!!!!!!!

Advocard [28]

Answer:

KOH

Explanation:

Chemical reaction:

2K + 2H₂O → 2KOH + H₂

Element ratio of K.

K = 1

Because only potassium is present.

Element ratio of H₂O.

2 : 1

in water ratio of element is 2 : 1 because two hydrogen and one oxygen atom present.

Element ratio of KOH

1 : 1 : 1

in KOH elemental ratio is 1 : 1 : 1 because one potassium one hydrogen and one oxygen atom are present.

Element ratio of H₂.

2

Just two atoms of hydrogen are present.

8 0

3 years ago