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Varvara68 [4.7K]
3 years ago
14

Will mark brainliest

Chemistry
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer:

4. Principal and Azimuthal (subsidiary) quantum number

5.Principal, Azimuthal (subsidiary), and magnetic quantum number

6. 10 electrons

7. 32 electrons

8. 36 electrons

Explanation:

4. Principal and Azimuthal (subsidiary) quantum number because in 4d, 4 represent principal quantum number and d- represents azimuthal quantum number (having l- value as 3)

5.Principal, Azimuthal (subsidiary), and magnetic quantum number are the first three because 2 stands for principal, s-for azimuthal (l=0) and magnetic quantum number for s- orbital= 0

6. 10 electrons, because for sublevel with l= 3, is a d-sub-level, and d- can take 10-electrons

7. 32 electrons, using the relationship 2×n^2 for the maximum number of electrons in a shell,

,n= 4 , hence 2×4^2= 32

8. 36 electrons, because n=4 and n= 3 can have the maximum configuration of [Ar]4s^2 3d^10 4p^6

This will sum up to 36- electrons, since Argon has 18 -electrons.

18+2+10+6=36 electrons

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Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answ
inessss [21]

Answer:

\Delta S^{0} for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: \frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)

\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]

where S^{0} represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K

So, \Delta S^{0} for the given reaction is -99.4 J/K

7 0
3 years ago
2. A golden-colored cube is handed to you. The person wants you to buy it for
lara31 [8.8K]

Answer:

Volume of the cube is 2*2*2 = 8cm^3.

Divide the weight (40g) over the volume (8cm^3).

Density is 5g/cm^3. Definitely not gold.

Explanation:

4 0
3 years ago
Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with m
daser333 [38]

The question is incomplete, the complete question is;

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with multiple compounds.

Experimental melting point is BELOW literature value

Experimental melting point is CLOSE to literature value

WIDE melting point range

NARROW melting point range

Answer:

narrow melting point-pure sample of a single compound

experimental melting point is close to literature value-pure sample of a single compound

wide melting point range-impure sample of multiple compounds

experimental melting point is below literature value-impure sample of multiple compounds

Explanation:

The experimental melting point of a pure single compound is sharp and extremely close to the melting point of the substance as recorded in the literature. Usually, a pure substance melts within a narrow range of temperatures.

Impure samples of multiple compounds melt over a range of temperatures. Also if the experimental melting point is well below the record in literature, then the sample is contaminated by other compounds.

4 0
3 years ago
Help needed ASAP plz
Sati [7]

Answer:

6 cucumberslices.

Explanation:

You can make a maximum of 4 sandwiches. 8 slices of tomato and 2 needed for each. 8/4=2. You need 4 slices of cucumber per sandwich and have 22. the maximum of sandwiches we can make is 4 so 4×4=16. 22-16=6 leftover slices of cucumber.

7 0
3 years ago
Determine the wavelength of the energy that needs to be absorbed for a 3p electron in chlorine to be promoted to the 4s subshell
bazaltina [42]

Answer:

The wavelength of the energy that needs to be absorbed  = 52.36 nm

Explanation:

For this study;

Let consider the Rydgberg equation from Bohr's theory of atomic model:

i.e.

\dfrac{1}{\lambda} = R_H (Z^*)^2( \dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

where

Z* = effective nuclear charge of atom = Z - σ = 6

n₁ = lower orbit = 3

n₂ = higher orbit = 4

R_H = Rydyberg constant = 1.09 × 10⁷ m⁻¹

λ = wave length of the light absorbed

∴

\dfrac{1}{\lambda} = 1.09 \times 10^7}(6)^2( \dfrac{1}{3^2}-\dfrac{1}{4^2})

\dfrac{1}{\lambda} = 1.09 \times 10^7}(36)( \dfrac{1}{9}-\dfrac{1}{16})

\dfrac{1}{\lambda} = 392400000\times0.0486111111

\dfrac{1}{\lambda} =19075000

\lambda = \dfrac{1}{19075000}

\lambda = \dfrac{1}{1.91\times 10^7 \ m^{-1}}

\lambda = 5.236 \times 10^{-8} m

\lambda = 52.36 \times 10^{-9} m

\lambda = 52.36\  n m

Therefore, the wavelength of the energy that needs to be absorbed  = 52.36 nm

7 0
3 years ago
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