Posted by Lily on Tuesday, November 16, 2010 at 7:56pm.
What's the problem Lily?
how do you do it? like how do i write it like not in word but like with the symbols
For the first one.
H has a valence of +1; I has a valence of -1; therefore, the formula for hydrogen iodide is HI.
You know the symbol for hydrogen. It is H but when it is by itself it is diatomic so you write H2.
You know the symbol for iodine. It is I but when by itself it is diatomic or I2.
Note: There are 9 elements which, in their free state, are not monatomic. They are H2, N2, O2, F2, Cl2, Br2, I2, and their cousins S8 and P4. If you will look for their location in the periodic table, you will see how you can memorize what they are.
Second one.
Symbol for aluminum is Al.
Symbol for iodine is I2.
Formula for aluminum iodide is AlI3 (because Al has valence of +3 and I has a valence of -1.)
Must you memorize all of the valences? A few, yes, but most, no. Note Al is in group III (or 13 depending upon the system being used) and the valence is 3. I is in group VII (or 17) and the valence is -1. H is in group I (or 1) and has a valence of +1.
So for those elements in group I, II, III, valence is +1, +2 or +3 respectively. For those in group V, VI, VII, valence is -3, -2, -1 respectively
So the first one would be:
H2+I--> HI2?
No. The first one is.
H2 + I2 == HI
The idea behind writing the "skeleton" equation is to get the symbols and formulas down pat. You don't worry about balancing the equation yet. Balancing comes later.
okay. if i was balancing it though would it be H2+ I2-->2HI?
and for the second one to see if im doing this right:
Al3+I2--> AlI3?
Almost.
Al + I2 ==> AlI3 is the skeleton equation. Al is not one of those I listed a couple responses ago that is not monatomic. Just H2, N2, O2, F2, Cl2, Br, I2, P4, S8 are not monatomic (1 atom to the molecule).
Yes, H2 + I2 ==> 2HI is the balanced equation for the formation of HI from elements.
So the reactants dont need the charge number only the products
but for letter c. what do you do when you have two elements would it be FeO3?
The nature of the substance is what determines how the formula is written, not whether it is a reactant or a product. In the first one, hydrogen is written H2 because it is one of those 9 elements I mentioned AND it is in the free state. I2 is the same thing. The formula for hydrogen iodide is HI because that is a COMPOUND and elements combine in certain proportions when they make compounds. Since the valence of H is +1 and that of I is -1, the formula is HI instead of some other set of numbers.
For c, we have
c. iron(II) oxide(s) + oxygen(g) --> iron(III) oxide(s)
iron(II) oxide is a compound. Iron is one of those elements that may have more than one valence and the II tells us it is +2 in this compound. Since the valence of oxygen is -2, the formula for iron(II) oxide is FeO. That plus oxygen (oxygen is in the free state so we must write it as O2). (s) for FeO means it is a solid, g beside O2 means O2 is a gas. iron(III) oxide is Fe with a valence of +3 and oxide is oxygen with a valence of -2, we put them together to make a compound of iron(III) oxide with the formula Fe2O3. Note that the valences add to zero in compounds. Fe is +3 x 2 = 6 of + charges. Oxygen is -2 x 3 = -6 charges. and +6 with -6 = 0. That is one way to do it when the numbers don't come out even (as they do is NaCl, MgO, etc). Here is a very easy way. I can't draw the diagram on the board but I'll explain it.
DRAW on a sheet of paper a large X between Fe and O something like this.
FeXO. Above the Fe write its valence of +3. Above the O write its valence of -2. You will notice that the +3 is at the top of one of the legs of the X. The -2 is at the top of the other leg of the X. Just follow the +3 leg down and the 3 goes as the subscript for O. Follow the -2 down and it becomes the subscript for the Fe (without the - sign, of course). It works every time BUT you must remember to reduce the number if possible. For example Mg and O come out to be Mg2O2 so you must remember to divide by 2 to make it MgO. When writing the formulas, you should place that X mentally and not actually in writing.
okay now i get it. thanks so much
sorry but i want to make sure im doing this right. for lithium+gold(III chloride-->lithium chloride+gold
would it be Li+AuCl2-->LiCl2+Au?
sorry but i want to make sure im doing this right. for lithium+gold(III chloride-->lithium chloride+gold
would it be Li+AuCl2-->LiCl2+Au?
gold(III) chloride means Au^+3 and Cl^- so that will be AuCl3 for the formula. Li is +1 and Cl is -1 so lithium chloride is LiCl.
okay im sure i can get one right soon.
now if it has a polyatomic in the equation like Tin(IV) nitrate would it be SnNO3? and if it produces iron(III)nitrate would it be Fe(NO3)3
Iron(III) nitrate is Fe(NO3)3.
Sn(IV) nitrate is Sn(NO3)4
okay im getting closer(:
so would nicket(II) oxide be NiO2
and dichlrine pentoxide be Cl2O5
dichloropentoxide is Cl2O5.
nickel(II) oxide is NiO. Ni is +2, O is -2.
If you have other questions, please start a new post at the top of the first page. This one takes too long to check to see if you have other issues.
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