3.25

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sandros
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

3.25

Postby sandros » Sun Nov 05, 2017 7:43 pm

Can someone please explain the process ? I can't figure out the correct answers for that one ...

Sabah Islam 1G
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: 3.25

Postby Sabah Islam 1G » Sun Nov 05, 2017 8:07 pm

To find the chemical formula of the compounds, just take the value of the charge of one atom and make it the subscript of the other atom in the compound. For example, when writing the chemical formula for magnesium arsenide, the charge for Mg (magnesium) is +2, so 2 would be the subscript for As (arsenic), and the charge for As is -3, so 3 would be the subscript for Mg. Therefore, the chemical formula for magnesium arsenide would be Mg3As2.

Timothy Kim 1B
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: 3.25

Postby Timothy Kim 1B » Sun Nov 05, 2017 10:40 pm

With these two charges, find the lowest common factor between the two. Use this lowest common factor as the subscript for the opposite charges.

Dylan Mai 1D
Posts: 70
Joined: Sat Jul 22, 2017 3:00 am

Re: 3.25

Postby Dylan Mai 1D » Tue Nov 07, 2017 6:44 pm

How do you know the charges for the elements? memorization?

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

Re: 3.25

Postby Aijun Zhang 1D » Wed Nov 08, 2017 10:05 pm

Dylan Mai 1D wrote:How do you know the charges for the elements? memorization?

It's basically based on electron configuration, or say, how it would be most stable to form a compound.
For example, for group 1 and 2, it is most possible that they are tend to lose electrons when forming a compound. Then the number of electrons lost is the charge they would have. Li only has 1 valence electron, so in a compound, it's oxidation number will be +1.
For group 14 to 17, they are more likely to gain electrons to fullfill their orbitals. So for oxygen, it usually gains 2 electrons and thus have an oxidation number of -2. But it is a little bit complicated when it comes to N, S, P and C (and sometimes O).
For d-block elements, one of the property is that they have various oxidation numbers. So it is much harder.

For what I remembered, the actual way to calculate the oxidation number is to use formal charge. But I cannot recall very well now.
Generally I would recommend to remember some of the oxidation numbers of common elements, such as H, Na, O(usually-2, sometimes -4 and -1), Cl, etc. Then use the unchanged one to calculate the one has various oxidation states.

RenuChepuru1L
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

Re: 3.25

Postby RenuChepuru1L » Thu Nov 09, 2017 10:33 am

what happens when there is a number in parentheses for example Bismuth(III)?

Hammad Khan 2B
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am
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Re: 3.25

Postby Hammad Khan 2B » Thu Nov 09, 2017 10:43 am

RenuChepuru1G wrote:what happens when there is a number in parentheses for example Bismuth(III)?


The number in the parentheses just means the charge of the element. Ex) Cu (II) would be Cu2+ and so on.


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