So what you’re going to do is basically the + and - in each top hand corner is the charge of compound, so for example Li has a charge of +1 while Br has a charge of -1 , to write the formula you need to get the charges to cancel out ( equal zero) so luckily this was easy because -1 +1 =0 ! So it would be LiBr. Though for another example Al has a charge of 3+ while br has a charge of -1 and these do not equal zero, so as a result you have to add more br making the Formula AlBr3! Hope this helps!
Answer:
2, 1, 1, 4.
Explanation:
Hello there!
In this case, for the given chemical reaction:

We can see how there is one SO4 on the left and two on the right, thus, we add a 2 in front of H2SO4:

Next, since there are 8 atoms of hydrogen on the left and two on the right, we add a 4 in front of H2O to obtain:

Which is now balanced so the coefficients 2, 1, 1, 4.
Best regards!
Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions. The “ingredients” of a reaction are called reactants, and the end results are called products.
Is there choices to this question? cant answer it without choices
Answer : The concentration of
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is
and
is excess reagent.
First we have to calculate the moles of KSCN.


Moles of KSCN = Moles of
= Moles of
= 
Now we have to calculate the concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of
is, 