I have a question. I may just be preparing my equation wrong perhaps however, can an element have two different charges on both sides on the equation? for example, can Nickel have a 2+ charge on the left of the equation and say a 3+ on the right?
If you're doing double displacement reactions then the charge will not change between reactans (left) and products (right). Charges only change in what are known as oxidation-reduction or redox reactions.
Thank you very much! I am learning redox reactions now also and they can be tricky. I havent looked ye, but do you have videos on those too? Thanks again for all your time and help!
I have a question.. how are the hydrogens on the last rxn equal. the reactant side has "2H" and the product side has "2H2" I thought that equals 4 H on the product side.. or is it because Hydrogen is diatomic?
The reactants side is Ba(OH)2 + 2 HI. There are 2 H in Ba(OH)2 and 2 more H in 2HI for a total of 4 H. On the products, there are 2 H2O which is also a total of 4 H.
Best lesson on the subject on youtube, thank you so much for your time and energy!
Great video, thanks!
Thanks for the encouragement! Be sure to tell your friends. :)
I have a question. I may just be preparing my equation wrong perhaps however, can an element have two different charges on both sides on the equation? for example, can Nickel have a 2+ charge on the left of the equation and say a 3+ on the right?
If you're doing double displacement reactions then the charge will not change between reactans (left) and products (right). Charges only change in what are known as oxidation-reduction or redox reactions.
Thank you very much! I am learning redox reactions now also and they can be tricky. I havent looked ye, but do you have videos on those too? Thanks again for all your time and help!
I have a question.. how are the hydrogens on the last rxn equal. the reactant side has "2H" and the product side has "2H2" I thought that equals 4 H on the product side.. or is it because Hydrogen is diatomic?
The reactants side is Ba(OH)2 + 2 HI. There are 2 H in Ba(OH)2 and 2 more H in 2HI for a total of 4 H. On the products, there are 2 H2O which is also a total of 4 H.