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olga2289 [7]
3 years ago
7

Barium (Ba) has two valence electrons, and these electrons are located in the 6s subshell. Without using the periodic table, in

which group and period is barium located?
Group 2A, Period 7

Group 1A, Period 6

Group 1A, Period 7

Group 2A, Period 6
Chemistry
2 answers:
Katena32 [7]3 years ago
7 0
Barium is located in Group 2A, Period 6
artcher [175]3 years ago
5 0

Answer:

Group 2A, Period 6

Explanation:

To know where an atom is located, we need to see its electronic distribution. The major number in it will be the period (not necessary is the number of the valence, because subshell 4s is less energetic than 3d, so it will occur first!).

To know the group, we need to check the subshell more energetic, if is subshell <em>s</em>, it will be in group 1A or 2A, if it has only 1 valence electron, then its group 1A, with two valence electrons, is group 2A.

If the subshell more energetic is the subshell <em>p</em>, so it will be from 3A to 8A, we just add the number of electrons of subshell <em>p</em> with the numbers of electrons of subshell <em>s</em> of the same layer, and we'll have the number of the group.

If the subshell more energetic is the subshell <em>d</em>, it will be in group B, and the number of electrons will be the number of the group.

If the subshell more energetic is subshell <em>f</em>, then it is on group 3B, because will be in latanids or actinides.

So, Barium has the subshell 6s with 2 valence electrons, then it is in group 2A, Period 6.

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Find the reagent to sythesize 1- phenyl-1-propanol
vova2212 [387]

Answer:

aqueous acid is used as a reagent.  

Explanation:

Addition of Grignard reagent in aldehyde and followed by the acidification give rise to the primary or secondary alcohol. when the formaldehyde is used than the primary alcohol is formed otherwise secondary alcohol is formed.

in this reaction we also use the aqueous acid for the acidification as a reagent. We add aqueous acid when ethanol is present. This is because ethanol is get converted in the presence of aqueous acid into the chloroethane.  

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2 years ago
If Tommy had started with 0.75 moles of potassium carbonate how many moles of
worty [1.4K]

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Explanation: suggest watching a video

4 0
3 years ago
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4 0
2 years ago
I NEED HELP PLEASE! :)
MrMuchimi

<u>answer</u> 1<u> </u><u>:</u>

Law of conservation of momentum states that

For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

<u>answer</u><u> </u><u>2</u><u>:</u><u> </u>

When a substance is provided energy<u> </u>in the form of heat, it's temperature increases. The extent of temperature increase is determined by the heat capacity of the substance. The larger the heat capacity of a substance, the more energy is required to raise its temperature.

When a substance undergoes a FIRST ORDER phase change, its temperature remains constant as long as the phase change remains incomplete. When ice at -10 degrees C is heated, its temperature rises until it reaches 0 degrees C. At that temperature, it starts melting and solid water is converted to liquid water. During this time, all the heat energy provided to the system is USED UP in the process of converting solid to the liquid. Only when all the solid is converted, is the heat used to raise the temperature of the liquid.

This is what results in the flat part of the freezing/melting of condensation/boiling curve. In this flat region, the heat capacity of the substance is infinite. This is the famous "divergence" of the heat capacity during a first order phase transition.

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4 0
3 years ago
On the basis of the information above, a buffer with a pH = 9 can best be made by using
telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for \mathbf{H_3PO_4 , H_2PO^{-}_4 ,  HPO_4^{2-}} are \mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \  5\times 10^{-13}} respectively.

\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}

\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}

\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}

The reason while option D is the best answer is that, the value of pKa for both

\mathbf{H_2PO^{-}_4 ,\  \& \  HPO_4^{2-}} lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}

3 0
2 years ago
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