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Marrrta [24]
2 years ago
14

What species has an atom with an oxidation number of 3

Chemistry
1 answer:
Montano1993 [528]2 years ago
4 0
So we found out that chlorine had an overall oxidation number of plus three, and that's it for this one.
You might be interested in
The number of electrons in n=1 and n=2 shells of aluminum are
harina [27]

Answer:

n=1 holds two electrons and n=2 holds eight electrons.

Explanation:

Hello

In this case, since the atomic number of aluminum is 13, its electron configuration is:

Al^{13}: 1s^2,2s^2,2p^6,3s^2,3p^1

In such a way, we can see that the level n=1 is filled with two electrons since the subshell s is able to hold two electrons and the level n=2 is also filled but with eight electrons as s holds two whereas p holds 6. Moreover, n=3 is holding three electrons.

Best regards.

5 0
3 years ago
How do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved
IRINA_888 [86]

Answer:

Molar mass of MgCl2 is 95 g/mol

Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol

moles = mass given/ molar mass

= 2.7/95 = 0.028 mol

volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)

molarity of MgCl2 = moles/volume

= 0.028/0.25

= 0.112 mol/dm3

3 0
2 years ago
Which of these compounds is not a molecule?<br><br> A) MgO<br><br> B)SO2<br><br> C)CO<br><br> D) CS2
DerKrebs [107]
A) MgO
Hope that helps
5 0
3 years ago
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Katena32 [7]

Answer: Uhm you answered your own question...

Explanation:

7 0
3 years ago
Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm
pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0
3 years ago
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