Answer:
See attached figure.
Explanation:
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In this case, according to the given substances, we recall the concept of Lewis structure as such showing the bonds and valence electrons each atom has in the molecule. Thus, since chlorine atoms have seven valence electrons, carbon atoms four of them, hydrogen atoms have 1 and oxygen atoms 6, we are able to draw such Lewis dot structures, by obeying the octet as shown on the attached figure.
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Periodic table is arranged and organized with special pattern or regular variation of the properties of an element with increasing atomic number, this is called periodic trend. So periodic trend look like a repeating pattern on the periodic table.
Some of the trends are density, atomic radius,melting point, boiling point, electronegativity etc.
We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.
Dividing the mass of each reactant by its molar mass:
(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6
(10 g O2)(31.999 g/mol) = 0.3125 mol O2.
Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.
Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).
So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.
Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.
Answer:
Atom - un átomo de hidrógeno pierde un electrón, adquiere una carga positiva
Molecule - Al aumentar el calor, las moléculas se alejan en mayor medida unas de otras y se intensifica su estado de agitación.
Answer:The 1st and 2nd reactions are the example of oxidation -reduction.
Explanation:
Oxidation is basically when a species loses electrons and reduction is basically when the species gains electrons.
A reaction is known as an oxidation -reduction reaction only if oxidation and reduction simultaneously occur in the reaction. It basically means if a species is getting oxidized in the reaction then the other species present in the system must be reduced in the reaction.
Oxidation-reduction reactions are also known as redox reactions.
In the 1st reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of chlorine is 0 in reactants and in products is -1 so chlorine is reduced. Hence Na is oxidized and Cl is reduced so the reaction is a example of oxidation-reduction.
2Na(s)+Cl₂(g)→2NaCl(s)
In the second reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of Cu is +1 in reactant and 0 in products so Cu is reduced. Hence Na is oxidized and Cu is reduced so the reaction is an example of oxidation-reduction.
Na(s)+CuCl(aq)→NaCl(aq)+Cu(s)
In the third reaction the oxidation state of Na changes from +1 to +1 and that of Cu also changes from +1 to +1. So there is no change in oxidation state of the species present in reactants and products. Hence this reaction is not an example of oxidation and reduction.
