Answer:
Resonance hybrid: A weighted average of all significant resonance contributors depicting the true electronic structure of a molecule.
Explanation:
Translation: Híbrido de ressonância: uma média ponderada de todos os contribuintes de ressonância significativos, representando a verdadeira estrutura eletrônica de uma molécula.
Answer:
Hydrogen
Explanation:
A reducing agent is a substance which gives up its electrons to become oxidized. Generally, metals are oxidized (reducing agents) while non-metals are reduced (oxidizing agents).
However, hydrogen which is a non-metal is usually oxidized in the presence of stronger oxidizing non-metals such as fluorine and oxygen.
Hydrogen thus, acts as a reducing agent by giving up its electrons to become oxidized. Even though among all non-metals, Hydrogen has the greatest potential to be oxidized, it is a poor reducing agent compared to reactive metals.
Answer:
Explanation:
Mass of one tablet = 20 mg
Mass of two tablets = 
Percent that is soluble in water = 40%
Mass of tablet that is soluble in water = 
So, mass of solute is 
Density of water = 1 kg/L
Volume of water = 1 L
So, mass of 1 L of water is 
PPM is given by
Hence, the concentration of iodine in the treated water .
The amount of water that will be produced is 50.36 grams
<h3>Stoichiometric problems</h3>
The metabolism of glucose is represented by the following equation:
The mole ratio of glucose metabolized to the water produced is 1:6.
Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles
Equivalent mole of water = 0.4662 x 6 = 2.7975 moles
Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams
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39.25 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂ + O₂ → 2 H₂O
Now we calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of H₂ = 14.8 / 2 = 7.4 moles
number of moles of O₂ = 34.8 / 32 = 1.09 moles
We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:
if 1 moles of O₂ produces 2 moles of H₂O
then 1.09 moles of O₂ produces X moles of H₂O
X = (1.09 × 2) / 1 = 2.18 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 2.18 × 18 = 39.25 g
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limiting reactant
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