The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction, which requires that species exhibits changing oxidation statesduring the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components and is often separated into independent two hypothetical <span>half-reactions </span>to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:
<span><span><span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span>(1)</span><span>(1)<span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span></span></span>
The first step in determining whether the reaction is a redox reaction is to splitting the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:
<span><span><span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span>(2a)</span><span>(2a)<span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span></span></span>
The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in Cu to +2 in Cu2+. Now consider the silver atoms
<span><span><span>2A<span>g+</span>(aq)→2Ag(s)</span>(2b)</span><span>(2b)<span>2A<span>g+</span>(aq)→2Ag(s)</span></span></span>
In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an element on its own. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.
Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.
The periodic table of the elements are describe the electronic configuration of the elements on which the properties of the elements depends. Among the given groups only metal, non-metal and semi-metal group are the part of periodic table. The metallic property depends upon the binding energy of the electrons with the nucleus. Thus the elements which have the valence electrons more near to the nucleus that is s-block elements are more metallic in nature. On the other hand the elements which have the valence electrons far from the nucleus are more non-metallic in nature like p-block elements. However the binding energy or the attraction of the outermost electrons to the nucleus depends not only its valence electrons position but also some other factors like shielding effect, effective nuclear charge etc.
The elements which are in between the metals and non-metals can be classified as semi-metals.
Although the conductivity of a material is an inherent property of the metals but sometime the nonmetals or semi-metals are also behave like a conductor due to presence of the other elements, thus it cannot be a p[property of the periodic table. Similarly acidity, flammable gases are not part of the periodic table.
<h3>Answer:</h3>
4 mol O₂
<h3>General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 4 mol H₂O
[Solve] x mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol H₂O → 2 mol O₂
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Multiply/Divide:
Answer:
its 0.163 g
Explanation:
From the total pressure and the vapour pressure of water we can calculate the partial pressure of O2
PO 2 =P t −P H 2 O
= 760 − 22.4
= 737.6 mmHg
From the ideal gas equation we write.
W= RT/PVM = (0.0821Latm/Kmol)(273+24)K(0.974atm)(0.128L)(32.0g/mol/) =0.163g
