Answer:
N₂O₅ + H₂O --> 2HNO₃
This reaction is a combination reaction
Explanation:
I find that making a table is an easy way to balance a table. The table would be something like this:
Reactants (left) Products (right)
N = 2 N = 1
O = 6 O = 3
H = 2 H = 1
The number next the symbol represents how many of atoms of that particular element are present.
The aim is to make the number of atoms of each element on the left side to be equal to the number of atoms of each element on the right side. As we can see, there are 1 more nitrogen than on the right, 3 more oxygen than on the right and 1 more hydrogen than on the left.
So to make the numbers of atoms of each element on right equal to the number of atoms of each element on the left, we have to add a number. This number CANNOT be a subscript number because that would change the reaction.
We can add a 2 in front of the product (this is because there are less atoms in the right side of the equation).
N₂O₅ + H₂O --> 2HNO₃
This means there are now two HNO₃ molecules so every atom in this molecule is basically multiplied by 2. So 1 nitrogen atom becomes two (1 × 2 = 2), 3 oxygen atoms become 6 (3 × 2 = 6) and 1 hydrogen atom becomes 2 (1 × 2 = 2). If we were to make a table again with the following equation - N₂O₅ + H₂O --> 2HNO₃, the table would be as so:
Reactants (left) Products (right)
N = 2 N = 2
O = 6 O = 6
H = 2 H = 2
Now the equation is balanced as we can see the number each type of atom is the same on the right and left side.
Answer:
40 mL
Explanation:
Volume=400g/10(g/mL)
= 40mL
hope this will help you ❤️
I am 100% positive it is C. Have a ice day ( ^ω^ )
Baking soda acts as a base when exposed to acid in the stomach. This helps people who suffer from indigestion
Oxidation means that atom or ion loses electrons.
Reduction means that atom or ion takes electrons.
Cu²⁺ +2e⁻ ---> Cu⁰ , We can see that ion Cu²⁺ takes electrons, so this is reduction.
Fe⁰ ------> Fe⁺³ + 3e⁻, We can see that atom Fe lost electrons, so this is oxidation.
Reduction:
A., C., E.
Oxidation:
B., D.
