Hydrogen peroxide, H2O2, decomposes according to the equation above. This reaction is thermodynamically favorable at room temper
ature.
(a) A particulate representation of the reactants is shown below in the box on the left. In the box below on the right, draw the particulate representation of all the molecules that would be produced from these four reactant molecules.
(a) A particulate representation of the reactants is shown below in the box on the left. In the box below on the right, draw the particulate representation of all the molecules that would be produced from these four reactant molecules.
1 answer:
Answer:
4H₂O₂ → 4H₂O + 2O₂
Explanation:
The chemical reaction for the decomposition of hydrogen peroxide can be represented as follows;
2H₂O₂ → 2H₂O + O₂
Therefore, two molecules of hydrogen peroxide decomposes into two molecules of water and one molecule of oxygen.
In a related diagram, we have 4 molecules of hydrogen peroxide decomposing, therefore we have;
4H₂O₂ → 4H₂O + 2O₂
The attached diagram shows the products of the decomposition of the four molecules of hydrogen peroxide.
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Answer:
See explanation below
Explanation:
In this case, let's see both molecules per separate:
In the case of SeO₂ the central atom would be the Se. The Se has oxidation states of 2+, and 4+. In this molecule it's working with the 4+, while oxygen is working with the 2- state. Now, how do we know that Se is working with that state?, simply, let's do an equation for it. We know that this molecule has a formal charge of 0, so:
Se = x
O = -2
x + (-2)*2 = 0
x - 4 = 0
x = +4.
Therefore, Selenium is working with +4 state, the only way to bond this molecule is with a covalent bond, and in the case of the oxygen will be with double bond. See picture below.
In the case of CO₂ happens something similar. Carbon is working with +4 state, so in order to stabilize the charges, it has to be bonded with double bonds with both oxygens. The picture below shows.
