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2 years ago

Give the quantum number set for one electron in the 3p sublevel of a sulfur (S) atom.

1 answer:

7 0

<span>sulfurs electronic configuration is 1s2 2s2 2p6 3s2 3p4 so for the 3p sublevel principal quantum no - 3 angulum quantum no - 1 magnetic quantum no - (-1) or 0 or (-1) spin quantum no - (+1/2) or (-1/2) so the quantum numbers will be [ 3 , 1, -1 , +1/2 ] or [ 3 , 1, -1 , -1/2 ] [ 3 , 1, 0 , +1/2 ] or [ 3 , 1, 0 , -1/2 ] [ 3 , 1, 1 , +1/2 ] or [ 3 , 1, 1 , -1/2 ]
Hope This Helps</span>

You might be interested in

Initial alcohol is converted to compound A. Compound A undergoes three different transformations. Name compound A and each the f

nikklg [1K]

<h2>Acetic Acid</h2>

Explanation:

Initial alcohol that is formed from methane is methanol that can be converted to acetic acid.

Methanol in presence of $CO_2$ and hydrogen gas gets oxidized to acetic acid with the release of water.

Hence, compound A will be Acetic acid.

$CH_3OH + CO_2 +H_2$ → $CH_3COOH + H_2O$

The acetic acid formed can be transformed into -

$CH_3COOH$ → $CH_4 + CO$

The product formed is methane and carbon monoxide.

2. $CH_3COOH$ → $CH_2CHO + H_2O$

The product formed is formaldehyde and water.

3. $CH_3COOH + NaHCO_3$ → $CH_3COONa +CO_2 +H_2O$

The product formed is sodium acetate, carbon dioxide, and water.

4 0

2 years ago

In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0

sp2606 [1]

Answer:

The value of $K_p$ for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = $p^o_{H_2O}=15.0 Torr$

Partial pressure of hydrogen gas at equilibrium = $p^o_{H_2}=?$

Total pressure of the system at equilibrium P = 36.3 Torr

Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

$P=p^o_{H_2O}+p^o_{H_2}$

$p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr$

$3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)$

The expression of $K_p$ is given by:

$K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}$

$K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066$

The value of $K_p$ for this reaction at 1200 K is 4.066.

6 0

3 years ago

Calculate the number of moles in 7.41 X 1017 atoms Ag.

jolli1 [7]

Answer:

1.23x10^-6 mole

Explanation:

A clear understanding of Avogadro's hypothesis proved that 1mole of any substance contains 6.02x10^23 atoms. This indicates that 1mole of Ag contains 6.02x10^23 atoms.

Now, 1f 1mole of Ag contains 6.02x10^23 atoms, then Xmol of Ag will contain 7.41x10^17 atoms i.e

Xmol of Ag = 7.41x10^17/6.02x10^23 = 1.23x10^-6 mole

4 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

2 years ago