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2 years ago

How many of the following species are paramagnetic? sc3+ br- mg2+ se?

1 answer:

8 0

Answer:
One: <u>Selenium</u> is Paramagnetic

Explanation:
Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.

Electronic configuration of Scandium;

Sc = 21 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹

Sc³⁺ = 1s², 2s², 2p⁶, 3s², 3p⁶

Hence in Sc³⁺ there is no unpaired electron.

Electronic configuration of Bromine;

Br = 35 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵

Br⁻ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶

Hence in Br⁻ there is no unpaired electron.

Electronic configuration of Magnesium;

Mg = 12 = 1s², 2s², 2p⁶, 3s²

Mg²⁺ = 1s², 2s², 2p⁶

Hence in Mg²⁺ there is no unpaired electron.

Electronic configuration of selenium;

Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴

Or,

Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹

Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.

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The Lyman series results from excited state hydrogen atoms transiting to

Nutka1998 [239]

Answer:

I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.

n = 5 4th excited state

n = 4 3rd excited state

n = 3 2nd excited state

n = 2 1st excited state

n = 1 ground state

Here are the possible spectral lines.

n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.

n = 4 to 3, 4 to 2, 4 to 1 = 3 lines

n = 3 to 2, 3 to 1 = 2 lines

n = 2 to 1 = 1 line. Add 'em up. I get 10.

b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.

c.The E for any level is -21.8E-19 Joules/n^2

To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.

So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.

Explanation:

8 0

2 years ago

How do I find the slope of this graph? Is it even possible?

nasty-shy [4]

Answer:

Not exactly But you can take the slope of the curved portion and the slope of the flatline.

It wont do you much good since your working for absorbance but if you ever see something like a temperature change you can use the slope(s) to find freezing points/melting

Explanation:

If you need to submit a slope you could use a best fit which is just point to point or you could break it up like i mentioned

4 0

2 years ago

During prophase, which of the following events

alexdok [17]

Answer:

Spindle fibers form.

DNA condenses into chromosomes.

Explanation:

Prophase is the first stage of mitosis. Mitosis produces 2 daughter cells from a parental cell.

Spindle fibers form - this is true. During prophase, the mitotic spindle forms. Later during mitosis the spindle attaches to the centromere of chromosomes and pulls them to opposite ends of the cell prior to division

DNA condenses into chromosomes. - this is true. In the nucleus during interphase, the DNA is relatively loosely compacted. However, prior to division, the DNA is condensed into structures called chromosomes which are then paired up and distributed to the daughter cell.

Chromosomes move to opposite ends of the cell. - this is false. This happens during anaphase

Nuclear membrane begins to re-form. - this is false. This happens during telophase.

7 0

3 years ago

Salicylamide can undergo an iodination by electrophilic aromatic substitution. Arrange the procedural steps in order to iodinate

Tanzania [10]

The procedural steps in order to get iodinate salicylamide are as follows:

It begins first at a laboratory.
Secondly, one has to dissolve salicylamide in ethanol.
Then one has to add sodium iodide.
Later on, one has to add sodium hypochlorite to the ice cold solution of salicylamide and sodium iodide.
Thereafter one has to sodium thiosulfate.
It is better to Acidify by a person adding about 10% of HCl.
The one has to collect the crude product through the use of vacuum filtration.
Lastly one then Recrystallize from hot ethanol.

<h3>What is Electrophilic aromatic substitution?</h3>

The Electrophilic aromatic substitution is known to be a kind of an organic reaction where an atom is said to be added or attached to a kind of aromatic structure (that is made up of hydrogen) is said to be replaced by an electrophile.

Learn more about electrophilic aromatic substitution from

brainly.com/question/14908357

4 0

2 years ago

16.25 g of water at 54 C relaeases 402.7 J. What will be its final temp?

leonid [27]

Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?

Solving:

Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
$\Delta\:T = \frac{-402.7}{67.99}$
$\boxed{\Delta\:T \approx -5.92\:^0C}$

If: ΔT (T final - T initial) = ?
$-5.92^0 = T_{final} - 54^0$
$T_{final} = 54^0 - 5.92^0$
$\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark$

8 0

3 years ago