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Whitepunk [10]
3 years ago
8

The following do not represent valid ground-state electron configurations for an atom either because they violate the Pauli excl

usion principle or because orbitals are not filled in order of increasing energy. Indicate which of these two principles is violated in each example or whether both or neither are violated.
Part 1
1s22s23s2
A) The orbitals are not filled in order of increasing energy.
B) The Pauli exclusion principle is violated.
C) Orbitals are not filled in order of increasing energy and the Pauli exclusion principle is violated.
D) The ground-state electron configuration is valid.
Part 2
[Rn]7s26d4
A) The orbitals are not filled in order of increasing energy.
B) The Pauli exclusion principle is violated.
C) Orbitals are not filled in order of increasing energy and the Pauli exclusion principle is violated.
D) The ground-state electron configuration is valid.
Part 3
[Ne]3s23d5
A) The orbitals are not filled in order of increasing energy.
B) The Pauli exclusion principle is violated.
C) Orbitals are not filled in order of increasing energy and the Pauli exclusion principle is violated.
D) The ground-state electron configuration is valid.
Chemistry
2 answers:
svetoff [14.1K]3 years ago
7 0

Answer:

<em>For both cases the answer is C</em>

Explanation:  

We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.

For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.

Kipish [7]3 years ago
7 0

Answer:

for both parts the ans is c

hope helps you

have a great day

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Help. The answer to question 1 I didn't mean to click so please answer that as well.
Zepler [3.9K]
We have that energy=specific heat * change in temperature * mass. Thus, we have the final temperature (22) minus the initial temperature (55) to equal -33 as our change in temperature. Our specific heat is in J/g*C, so we're good with that because g stands for grams and the aluminium is measured in grams. As there are 10 grams of aluminum, we have
10*(-33)*0.902=-298 ish as our final temperature

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4 0
3 years ago
A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu
tatyana61 [14]

Answer:

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

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