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3 years ago

What is the total number of orbitals found in the second energy level? openstudy?

Chemistry

2 answers:

kkurt [141]3 years ago

4 0

Explanation:

Second energy level consists of one "s" sub-shell and a "p" sub-shell. So, in a "s" sub-shell there is only one orbital. Whereas in a "p" sub-shell there are 3 orbitals which are $p_{x}$ , $p_{y}$ , and $p_{z}$ .

Also, each orbital can contain a maximum of two electrons.

Therefore, we can conclude that in total there are four orbitals (one from s and three from p) found in the second energy level.

Setler79 [48]3 years ago

3 0

On the second shell there are two individual subshells:
The "s" subshell has only 1 orbital with max. two electrons spinning around; and the so-called "p" subshell has 3 orbitals with max. 6 electrons (2 on each!)
In total, there are four orbitals with 8 revolving electrons on the second shell.
Hope could help :)

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2 years ago

A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I

murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

6 0

3 years ago

Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 an

avanturin [10]

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L= $1000 cm^3$

Density of octane= d = $0.703 g/cm^3$

Mass of octane ,m= $d\times v=0.703 g/cm^3\times 1000 cm^3=703 g$

Moles of octane = $\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol$

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

$55\%=\frac{6.166 mol}{\text{Total moles}}\times 100$

Total moles = 11.212 mol

Moles of hexane :

$45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100$

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = $0.655 g/cm^3$

Volume of hexane = V

$V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L$

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

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3 years ago

When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b

bazaltina [42]

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

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Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
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4 0

3 years ago