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3 years ago

How many d and f orbitals total are present in the n = 4 shell of a hydrogen atom? Enter your answer as a whole number.

Chemistry

1 answer:

sashaice [31]3 years ago

4 0

Answer:

12 orbitals

Explanation:

The suborbital = d orbital.

n = 4

The orbital is 4d.

There are 5 degenerate orbitals of d orbitals. Thus m value, -2, -1, 0, +1, +2 .

Thus, number of orbitals = 5

The suborbital = f orbital.

n = 4

The orbital is 4f.

There are 7 degenerate orbitals of f orbitals. Thus m value, -3, -2, -1, 0, +1, +2, +3

Thus, number of orbitals = 7

Total number of orbitals = 7 + 5 = 12

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You have ice cubes in your freezer at home. The ice cubes melt into water. This type of change is called a... O a. nuclear chang

blagie [28]

B ) its a physical change

3 0

3 years ago

Paradichlorobenzene, C6H4Cl2, is a component of mothballs. A solution of 2.00 g in 22.5 g of cyclohexane boils at 82.39 ∘C. The

Rina8888 [55]

Answer:

2.79 °C/m

Explanation:

When a nonvolatile solute is dissolved in a pure solvent, the boiling point of the solvent increases. This property is called ebullioscopy. The temperature change (ΔT) can be calculated by:

ΔT = Kb*W*i

Where Kb is the ebullioscopy constant for the solvent, W is the molality and i is the van't Hoff factor.

W = m1/(M1*m2)

Where m1 is the mass of the solute (in g), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).

The van't Hoff factor represents the dissociation of the elements. For an organic molecule, we can approximate i = 1. Thus:

m1 = 2.00 g

M1 = 147 g/mol

m2 = 0.0225 kg

W = 2/(147*0.0225)

W = 0.6047 mol/kg

(82.39 - 80.70) = Kb*0.6047*1

0.6047Kb = 1.69

Kb = 2.79 °C/m

4 0

3 years ago

Be sure to answer all parts. In winemaking, the sugars in grapes undergo fermentation by yeast to yield CH3CH2OH (ethanol) and C

Ede4ka [16]

Answer:

a)- Fermentation = - 2 816 kJ/mol

Respiration = - 1409.2 kJ/mol

b)C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)

c) combustion per mol of sugar

Explanation:

The fermentation of sugar is given as follows:

C₆H₁₂O₆ (s) + 6O₂ (g) → 6O₂ (g) + 6H₂O (l)

ΔHrxn = ΔHformation (products) - ΔHformation (reactants)

= (6 mol × -393.5kJ/mol)+ (6mol × -285.8kJ/mol) - (1 mol × -1260kJ/mol + 6mol × 0)

= -2 816 kJ/mol

The heat of combustion of ethanol is given as follows:

C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)

Let's assume that 1 mol of ethanol is burnt. The heat of reaction, ΔHrxn is given by this equation:

ΔHrxn = ΔHformation (products) - ΔHformation (reactants)

= (2 mol× - 393.5kJ/mol) + ( 3mol × -285.8kJ/mol) - [ (1mol × -235 kJ/mol) + ( 3mol × 0.0000kJ)]

= -1644 - (-235.2)

= - 1409.2 kJ/mol

Therefore, the combustion of ethanol is exothermic. In other words, heat iis given off (this is signified by the negative sign)

b) Ethanol, like any other fuel, burns up to give water and carbon dioxide. This is based on the assumption that the combustion is complete and no side reactions take place. The combustion of ethanol is given as follows:

C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)

The physical states are given in the parentheses.

c) From the comparisons of the ΔHrxn, sugar produces more energy.

4 0

3 years ago

For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con

Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

$A\rightarrow B$

Expt [A] M [B] M Rate [M/s]

1 3.40 4.16 1.82*10^-4

2 4.59 4.16 3.32*10^-4

3. 3.40 5.46 1.82*10^-4

$Rate = k[A]^{x}[B]^{y}$

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

$\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2$

To find y:

Divide rate of expt 3 by expt 1

$\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0$

Therefore: x = 2, y = 0

$Rate = k[A]^{2}[B]^{0}$

To find k

Use rate for expt 1:

$k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1$

6 0

2 years ago

The wording of these questions are confusing how would you set this up ?

xxMikexx [17]

Answer:

8. 171074.8 mL

9. 3475 mL.

Explanation:

8. Determination of the volume of the diluted solution.

Initial Molarity (M₁) = 14 M

Initial volume (V₁) = 523 mL

Final Molarity (M₂) = 0.0428 M

Final volume (V₂) =?

Using the dilution formula, we can obtain the volume of the diluted solution as follow:

M₁V₁ = M₂V₂

14 × 523 = 0.0428 × V₂

7322 = 0.0428 × V₂

Divide both side by 0.0428

V₂ = 7322 / 0.0428

V₂ = 171074.8 mL

Therefore, the volume of the diluted solution is 171074.8 mL

9. Determination of the volume of water added.

We'll begin by calculating the final volume of the solution. This can be obtained as follow:

Initial Molarity (M₁) = 3.2 M

Initial volume (V₁) = 973 mL

Final Molarity (M₂) = 0.7 M

Final volume (V₂) =?

M₁V₁ = M₂V₂

3.2 × 973 = 0.7 × V₂

3113.6 = 0.7 × V₂

Divide both side by 0.7

V₂ = 3113.6 / 0.7

V₂ = 4448 mL

Thus, the final volume of the solution is 4448 mL

Finally, we shall determine the volume of water added. This can be obtained as follow:

Initial volume (V₁) = 973 mL

Final volume (V₂) = 4448 mL

Volume of water added =?

Volume of water added = V₂ – V₁

Volume of water added = 4448 – 973

Volume of water added = 3475 mL

7 0

2 years ago