A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is pr

Response:

V1 = 20.3L

Clarification:

P2 = 811.4Pa

V2 = 25.6L

P1 = 1023.6Pa

V1 =?

To answer this query, we will utilize Boyle's law, which states that the volume of a gas at constant temperature is inversely related to its pressure.

In mathematical terms,

V = k / P, where k = PV

The relationship can be defined as P1 × V1 = P2 × V2 = P3 × V3 =......=Pn × Vn

This simplifies to P1 × V1 = P2 × V2

Let’s rearrange for V1

V1 = (P2 × V2) / P1

Substituting values gives

V1 = (811.4 × 25.6) / 1023.6Pa

So, V1 = 20771.84 / 1023.6

This results in V1 = 20.29L, rounded to 20.3L

Answer:

189.2 KJ

Explanation:

Provided Data

light wavelength = 632.8 nm

Convert nm to meters

1 nm = 1 x 10⁻⁹

632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m

What is the energy of 1 mole of photons?

Solution

Used Formula

                     E = hc/λ

where

E = energy per photon

h = Planck's Constant

Planck's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light

speed of light = 3 × 10⁸ ms⁻¹

λ = wavelength of light

Insert values into the equation

                   E = hc/λ

                   E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)

                   E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)

                  E = 3.141 x 10⁻¹⁹J

3.141 x 10⁻¹⁹J indicates the energy for a single photon

Next, we need to determine the energy for 1 mole of photons

It is known that

1 mole contains 6.022 x10²³ photons

Consequently,

     Energy for one mole of photons = 3.141 x 10⁻¹⁹J x  6.022 x10²³

     Energy for one mole of photons = 1.89 x 10⁵ J

Now convert J to KJ

1000 J = 1 KJ

1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ

Thus,

the energy for one mole of photons is 189.2 KJ

Response: Below are the orbitals responsible for each bond identified in citric acid per the attachment.

Response 1) σ Bond a: Carbon uses and Oxygen employs .

Clarification: The sigma bonds are formed through the hybrid orbitals of carbon and oxygen. This occurs at the 'a' location in the citric acid structure.

Response 2) π Bond a: Both Carbon and Oxygen have π orbitals.

Clarification: The π-bond at position 'a' consists of interactions between the π orbitals of carbon and oxygen.

Response 3) Bond b: Oxygen and Hydrogen solely utilizes the S orbital.

Clarification: The bonding at position 'b' includes oxygen and hydrogen atoms, with hydrogen utilizing its S orbital.

Response 4) Bond c: Carbon is and Oxygen is also .

Clarification: The bonding process at position 'c' involves both carbon and oxygen atoms with their respective hybrid orbitals.

Response 5) Bond d: Carbon atom has and the second carbon has .

Clarification: In position 'd', the bond formed between carbon atoms is , utilizing orbitals that underwent hybridization which are .

Response 6) Bond e: C1 has O .

C2 has .

Clarification: The carbon that contains oxygen and a double bond utilizes hybridized orbitals; conversely, carbon at C2 employs hybridized orbitals in this bonding at position 'e'.