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Gnoma [55]
3 years ago
15

NEED HELP! what is a orbital . describe

Chemistry
1 answer:
olchik [2.2K]3 years ago
6 0

Answer:

In chemistry and quantum mechanics, an orbital is a mathematical function that describes the wave-like behavior of an electron, electron pair, or (less commonly) nucleons. An orbital can contain two electrons with paired spins and is often associated with a specific region of an atom.

Explanation:

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Công thức của A có dạng Ca(hco3)x có ptk là 162 tìm x
harina [27]

Answer:

j

Explanation:

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5 0
2 years ago
scientists often look for and attempt to explain patterns in nature. explain why patterns are important
Ronch [10]

patterns are important because they prove life has meaning ( or if you mean  

Scientifically  is is that the ability to recognize and create patterns help scientists make predictions based on there observations.

8 0
3 years ago
Using the balanced equation N2+O2=2NO, how many grams of NO can be produced when 25.0 grams of N react?
Oxana [17]

Answer:

53.55gNO

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:

=25.0gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNO}{1molN_2}*\frac{30.01 gNO}{1molNO}\\\\=53.55gNO

Best regards!

8 0
3 years ago
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
2 years ago
What happens to the particles in a substance as it freezes?
Ierofanga [76]

they move around and eventually come and join to make a solid

8 0
3 years ago
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