Which of the following compounds will experience dipole-dipole forces of attraction?

What information is needed to determine the energy of an electron in a many-electron atom? a) n. b) l. c) ml. d)ms

mel-nik [20]

Answer:

The correct answer is b.

Explanation:

The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.

In the case of atoms with more than one electron, the quantum number l also determines the sublevel of energy in which an orbital is found, within a certain energy level. The value of l is designated by the letters s, p, d, and f.

Have a nice day!

8 0

3 years ago

Read 2 more answers

Pls help me will mark brainliest, chemistry equilibrium!!!

Sergeeva-Olga [200]

Answer:

See attached => LeChatelier's Principle

Explanation:

LeChatelier's Principle => If a stress is applied to a chemical reaction, the reaction chemistry will shift away from the applied stress and establish a new equilibrium having new concentration values different from the original concentration values.

There are three (3) principle stress factors that will cause disturb a chemical reaction and cause it to shift to establish a new equilibrium. These are ...

=> concentration effects,

=> temperature effects, and

=> pressure-volume effects.

The attached notes explain in general terms how to evaluate a chemical reaction under an applied stress factor and determine direction of shift to establish a new equilibrium stability. Compare and apply to the worksheet problems.

_______

Answers to worksheet (compare to attached notes)

1a. => increasing wt to product side => tilts right => shifts left

1b. => removing wt from reactant side => tilts right => shifts left,

1c. => increasing wt to product side => tilts right => shifts left

2a. => tilts left (excess wt from Hg), shifts right,

2b. => Increase pressure shifts toward lower molar gas volume side (right) Note: apply only to gas phase substances, Hg is in liquid phase.

3a. exothermic rxn => cooling => tilts left, rxn shifts right

3b. endothermic rxn => cooling => tilts right, rxn shifts left

3c. exothermic rxn => cooling => tilts left, rxn shifts right

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don't forget => brainliest :-)

3 0

3 years ago

(3) A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3; Ka2=6.20

shtirl [24]

Answer:

The pH of this solution is 1,350

Explanation:

The phosphoric acid (H₃PO₄) has three acid dissociation constants:

HPO₄²⁻ ⇄ PO4³⁻ + H⁺ Kₐ₃ = 4,20x10⁻¹⁰ (1)

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸ (2)

H₃PO₄ ⇄ H₂PO4⁻ + H⁺ Kₐ₁ = 7,50x10⁻³ (3)

The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).

We need to know what is the concentration of H⁺ for calculate the pH.

The moles of H₂PO₄⁻ are:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

The moles of H⁺ are, in the same way:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

So:

H₃PO₄ ⇄ H₂PO4⁻ + H⁺ Kₐ₁ = 7,50x10⁻³ (3)

X mol ⇄ (1x10⁻³-X) mol + (1x10⁻³-X) mol (4)

The chemical equilibrium equation is:

Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]

So:

7,50x10⁻³ = (1x10⁻³-X)² / X

Solving the equation you will obtain:

X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0

Solving the quadratic formula you obtain two roots:

X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles

X = 1,065x10⁻⁴

So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol

The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)

Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M

pH is -log [H⁺]. So the obtained pH is 1,350

I hope it helps!

5 0

2 years ago

The pollen grains move in water randomly because of

polet [3.4K]

A the collisions between water particles and polllen grains

6 0

2 years ago

Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride a

AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

4 0

2 years ago