Gr 10 chem help uwu

[Maki]

Sulfur Dioxide is slowly oxidised in the atmosphere to form sulfur trioxide

I have to make a word + balanced equation for this
I put
SO2 + O2 -> SO3
???
Am I on the right track? IDK

[Ana]

Hm, the answer I get is YX3....

---------- Post added at 04:37 PM ---------- Previous post was at 04:37 PM ----------

I can explain the rational to that if you want...

[Lavendrae]

Element Y has 7 valence electrons. A full shell is 8 electrons. So Element Y needs 1 more electron to create a full valence shell.

Element X has 3 valence electrons. This means it has a full valence beneath the outer shell, and then it has 3 extra electrons in its outer shell. It is most likely to donate 3 of its electrons to become stable.

If Element Y needs 1 electron, three Y's will need 3 electrons. One X will provide 3 electrons necessary for the Ys. This is why the formula would be XY3.


I'm pretty sure that's the correct explanation. :)

[Maki]

Element Y has 7 valence electrons. A full shell is 8 electrons. So Element Y needs 1 more electron to create a full valence shell.

Element X has 3 valence electrons. This means it has a full valence beneath the outer shell, and then it has 3 extra electrons in its outer shell. It is most likely to donate 3 of its electrons to become stable.

If Element Y needs 1 electron, three Y's will need 3 electrons. One X will provide 3 electrons necessary for the Ys. This is why the formula would be XY3.


I'm pretty sure that's the correct explanation. :)

Oh that makes sense!

You must spread some Reputation around before giving it to Lavendrae again.
UGH
I'll rep you soon, I promise!

Mind explaining why Ana's explanation isn't right, then? It was along my same train of thought

[7:41:40 PM] Ana Banana:each element wants a full set of 8
X is going to look for 1 more electron
Y wants to give away it's extra 3 because it's easier to give away 3 than go looking for 5 more
[7:41:46 PM] Ana Banana: Y is going to meet the first X, and give away one. That makes the X happy with a full set of 8, but Y still has two more extra ones.
[7:42:18 PM | Edited 7:42:30 PM] Ana Banana: So it goes looking for 2 more Xs to give away the last two. In total it'll have met up with 3 different Xs
[7:42:28 PM] ky: you need 3 xs to make the y completely stable
[7:42:31 PM] ky: yeah haha

[Ana]

oh you're right xD
I mixed up the variables xD

---------- Post added at 05:16 PM ---------- Previous post was at 05:15 PM ----------

[11/4/13 4:39:44 PM] Ana Banana: mkay
[11/4/13 4:39:55 PM] Ana Banana: "X has 7 valence electrons, and element Y has 3"
[11/4/13 4:40:03 PM] ky: oh hmm it seems i didnt switch em up

^your original post had it the other way around.

[Maki]

oh you're right xD
I mixed up the variables xD

---------- Post added at 05:16 PM ---------- Previous post was at 05:15 PM ----------

[11/4/13 4:39:44 PM] Ana Banana: mkay
[11/4/13 4:39:55 PM] Ana Banana: "X has 7 valence electrons, and element Y has 3"
[11/4/13 4:40:03 PM] ky: oh hmm it seems i didnt switch em up

^your original post had it the other way around.



OHHH LOL
Sigh

Thanks you guys!

[Lavendrae]

Oh that makes sense!

You must spread some Reputation around before giving it to Lavendrae again.
UGH
I'll rep you soon, I promise!

Mind explaining why Ana's explanation isn't right, then? It was along my same train of thought

[7:41:40 PM] Ana Banana:each element wants a full set of 8
X is going to look for 1 more electron
Y wants to give away it's extra 3 because it's easier to give away 3 than go looking for 5 more
[7:41:46 PM] Ana Banana: Y is going to meet the first X, and give away one. That makes the X happy with a full set of 8, but Y still has two more extra ones.
[7:42:18 PM | Edited 7:42:30 PM] Ana Banana: So it goes looking for 2 more Xs to give away the last two. In total it'll have met up with 3 different Xs
[7:42:28 PM] ky: you need 3 xs to make the y completely stable
[7:42:31 PM] ky: yeah haha

Girl nbd, just hoping it makes sense to you. :)

[Maki]

BUUUMP NEED HALp

[Raj]

BUUUMP NEED HALp

hit up skype g

or post that problem here

[Jules]

What do you mean by 'make a word'?
If your equation is SO2 + O2 -> SO3 then to balance it you'd just make it 2SO2 + O2 -> 2SO3

[Poem]

Sulfur Dioxide is slowly oxidised in the atmosphere to form sulfur trioxide

I have to make a word + balanced equation for this
I put
SO2 + O2 -> SO3
???
Am I on the right track? IDK

Yup, just balance it like juliaxo said :)
Word equation would just be "sulfur dioxide reacts with oxygen to produce sulfur trioxide",

[Maki]

What do you mean by 'make a word'?
If your equation is SO2 + O2 -> SO3 then to balance it you'd just make it 2SO2 + O2 -> 2SO3


how did u get that?? how did u even know to put those coefficients. still confused :(
when i see stuff like this for example
reactants:
S- 3
products:
S- 4
i start freaking out bc there is no easy coefficient to put,,,

compared to this-
reactants:
S-2
products:
S- 4
bc for that ^,
i can just put a coefficient of 2 on the reactant side
its like easy multiplication
ja feel?

[Jules]

how did u get that?? how did u even know to put those coefficients. still confused :(

Because whatever is on the products side has to be equal to what is on the reactants side. So, your original equation was SO2 + O2 -> SO3, and you have 1 sulfur and 4 oxygen on the reactant side, but 1 sulfur and 3 oxygen on the products side. So by adding the coefficients and making it 2SO2 + O2 -> 2SO3, you now have 2 sulfur and 6 oxygen on the reactants side and the products side, so the equation is balanced.



when i see stuff like this for example
reactants:
S- 3
products:
S- 4
i start freaking out bc there is no easy coefficient to put,,,

compared to this-
reactants:
S-2
products:
S- 4
bc for that ^,
i can just put a coefficient of 2 on the reactant side
its like easy multiplication
ja feel?

If I had numbers like S 3 and 4 I just multiply by whatever will make both the same number. It takes a while to get used to all of it, but my teacher would give us really complicated combustion reactions to balance where your ending coefficients of the balanced equations are numbers like 17 and 51 and they take like 20 minutes to balance, it really is just guessing and checking half the time.

[Maki]

Because whatever is on the products side has to be equal to what is on the reactants side. So, your original equation was SO2 + O2 -> SO3, and you have 1 sulfur and 4 oxygen on the reactant side, but 1 sulfur and 3 oxygen on the products side. So by adding the coefficients and making it 2SO2 + O2 -> 2SO3, you now have 2 sulfur and 6 oxygen on the reactants side and the products side, so the equation is balanced.



If I had numbers like S 3 and 4 I just multiply by whatever will make both the same number. It takes a while to get used to all of it, but my teacher would give us really complicated combustion reactions to balance where your ending coefficients of the balanced equations are numbers like 17 and 51 and they take like 20 minutes to balance, it really is just guessing and checking half the time.

Ahh I see!
thanks :)

could you explain balancing
HCl + Ca(OH)2 -> CaCl2 + H2O?
the last thing i got before raging in a fit was
4HCl + Ca(OH)2 -> 5CaCl + 3H2O lol

[Jules]

Ahh I see!
thanks :)

could you explain balancing
HCl + Ca(OH)2 -> CaCl2 + H2O?
the last thing i got before raging in a fit was
4HCl + Ca(OH)2 -> 5CaCl + 3H2O lol

9810
This is how I balance, I make REP tables (reactants, elements, and products) so it is easier to see what you gave on each side and what you need