Concept explainers
(a)
Interpretation:
The oxidation number for the Nitrogen in given
Concept Introduction:
Oxidation:
Loss of electrons from an atom ion or molecule during a
Example
Here
Oxidation number:
It is the charge of an atom, provided if the compound is composed of ions. On oxidation the oxidation number will increase and on reduction the oxidation number will decrease. It can be also called as degree of oxidation.
Example:
`
Oxidation number of
Oxidation number of
Oxidation number of
Oxidation number of
Here we can see that the oxidation number of copper is decreased and the oxidation number of magnesium is increased.
(b)
Interpretation:
The oxidation number for the Nitrogen in given
Concept Introduction:
Oxidation
Loss of electrons from an atom ion or molecule during a chemical reaction is known as oxidation. Oxidation state of atom ion or molecule will increase in this process. In simple it is the addition of oxygen. Reduction is gaining of electrons.
Example
Here
Oxidation number:
It is the charge of an atom, provided if the compound is composed of ions. On oxidation the oxidation number will increase and on reduction the oxidation number will decrease. It can be also called as degree of oxidation.
Example:
`
Oxidation number of
Oxidation number of
Oxidation number of
Oxidation number of
Here we can see that the oxidation number of copper is decreased and the oxidation number of magnesium is increased.
(c)
Interpretation:
The oxidation number for the Nitrogen in given
Concept Introduction:
Oxidation
Loss of electrons from an atom ion or molecule during a chemical reaction is known as oxidation. Oxidation state of atom ion or molecule will increase in this process. In simple it is the addition of oxygen. Reduction is gaining of electrons.
Example
Here
Oxidation number:
It is the charge of an atom, provided if the compound is composed of ions. On oxidation the oxidation number will increase and on reduction the oxidation number will decrease. It can be also called as degree of oxidation.
Example:
`
Oxidation number of
Oxidation number of
Oxidation number of
Oxidation number of
Here we can see that the oxidation number of copper is decreased and the oxidation number of magnesium is increased.
(d)
Interpretation:
The oxidation number for the Nitrogen in given
Concept Introduction:
Oxidation:
Loss of electrons from an atom ion or molecule during a chemical reaction is known as oxidation. Oxidation state of atom ion or molecule will increase in this process. In simple it is the addition of oxygen. Reduction is gaining of electrons.
Example
Here
Oxidation number:
It is the charge of an atom, provided if the compound is composed of ions. On oxidation the oxidation number will increase and on reduction the oxidation number will decrease. It can be also called as degree of oxidation.
Example:
`
Oxidation number of
Oxidation number of
Oxidation number of
Oxidation number of
Here we can see that the oxidation number of copper is decreased and the oxidation number of magnesium is increased.
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Chapter 9 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- There is an instrument in Johnson 334 that measures total-reflectance x-ray fluorescence (TXRF) to do elemental analysis (i.e., determine what elements are present in a sample). A researcher is preparing a to measure calcium content in a series of well water samples by TXRF with an internal standard of vanadium (atomic symbol: V). She has prepared a series of standard solutions to ensure a linear instrument response over the expected Ca concentration range of 40-80 ppm. The concentrations of Ca and V (ppm) and the instrument response (peak area, arbitrary units) are shown below. Also included is a sample spectrum. Equation 1 describes the response factor, K, relating the analyte signal (SA) and the standard signal (SIS) to their respective concentrations (CA and CIS). Ca, ppm V, ppm SCa, arb. units SV, arb. units 20.0 10.0 14375.11 14261.02 40.0 10.0 36182.15 17997.10 60.0 10.0 39275.74 12988.01 80.0 10.0 57530.75 14268.54 100.0…arrow_forwardA mixture of 0.568 M H₂O, 0.438 M Cl₂O, and 0.710 M HClO are enclosed in a vessel at 25 °C. H₂O(g) + C₁₂O(g) = 2 HOCl(g) K = 0.0900 at 25°C с Calculate the equilibrium concentrations of each gas at 25 °C. [H₂O]= [C₁₂O]= [HOCI]= M Σ Marrow_forwardWhat units (if any) does the response factor (K) have? Does the response factor (K) depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)?arrow_forward
- Provide the structure, circle or draw, of the monomeric unit found in the biological polymeric materials given below. HO OH amylose OH OH 행 3 HO cellulose OH OH OH Ho HOarrow_forwardWhat units (if any) does K have? Does K depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)? in calculating the response factorarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
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