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3 years ago

Which types of orbitals are found in the principal energy level n = 2?

Chemistry

2 answers:

uranmaximum [27]3 years ago

5 0

Answer:

option d

Explanation:

plato users

AfilCa [17]3 years ago

3 0

Remember this:

1) n is principal quantum number and represents the energy level.

2) l is the second quantum number and represent the type of orbital.

3) l can take values from 0 to n - 1

4) each number of l is associated with a type of orbital. This table shows the equivalence:

l number       type of orbital

0                    s

1                    p

2                    d

3                     f

With that, you can tell that n = 2 permits l = 0 and 1, which is orbitals s and p.

Therefore, the answer is the option D) s, p.

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Protons + Neutrons = the atomic ?

kolezko [41]

Answer:

The atomic mass number is the number of protons and neutrons in an atom.

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3 years ago

How many liters of oxygen are in 8.32 moles of oxygen at STP

Jlenok [28]

Answer:

186 Liters at STP conditions

Explanation:

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3 years ago

The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall

miv72 [106K]

Answer:

The value is $C_p = 42. 8 J/K\cdot mol$

Explanation:

From the question we are told that

$\gamma = \frac{C_p }{C_v}$

The initial volume of the fluorocarbon gas is $V_1 = V$

The final volume of the fluorocarbon gas is $V_2 = 2V$

The initial temperature of the fluorocarbon gas is $T_1 = 298.15 K$

The final temperature of the fluorocarbon gas is $T_2 = 248.44 K$

The initial pressure is $P_1 = 202.94\ kPa$

The final pressure is $P_2 = 81.840\ kPa$

Generally the equation for adiabatically reversible expansion is mathematically represented as

$T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }$

Here R is the ideal gas constant with the value

$R = 8.314\ J/K \cdot mol$

$248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }$

=> $C_v = 31.54 J/K\cdot mol$

Generally adiabatic reversible expansion can also be mathematically expressed as

$P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$

=> $[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}$

=> $2^{\gamma} = 2.56$

=> $\gamma = 1.356$

$\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}$

=> $C_p = 42. 8 J/K\cdot mol$

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3 years ago

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Mice21 [21]

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denpristay [2]

C3H6 and CH4

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