The formula of the given compounds are to be written. Concept Information: Transition metals have the ability to form more than one positive ion and two ionic compounds with given non-metals. Stock system is the method for naming the metal ions that have more than one oxidation state in compounds indicated by Roman numeral in parenthesis just after the name of the metal ion. For naming of ionic compounds containing polyatomic ions, the name of the metal (cation) or a polyatomic cation written first, followed by the name of anion, this can be non-metal, and it this is non-metal (anion), change the suffix of non-metal to - i d e , adding at the end of the anion, that is in case of monatomic ions. The hydroxide ion and cynide ion are the exceptions. Rules of naming molecular compounds: Replace the ending of the second element with “ide”. While naming molecular compounds prefixes are used to indicate the number of a given element present in the compound like ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.
The formula of the given compounds are to be written. Concept Information: Transition metals have the ability to form more than one positive ion and two ionic compounds with given non-metals. Stock system is the method for naming the metal ions that have more than one oxidation state in compounds indicated by Roman numeral in parenthesis just after the name of the metal ion. For naming of ionic compounds containing polyatomic ions, the name of the metal (cation) or a polyatomic cation written first, followed by the name of anion, this can be non-metal, and it this is non-metal (anion), change the suffix of non-metal to - i d e , adding at the end of the anion, that is in case of monatomic ions. The hydroxide ion and cynide ion are the exceptions. Rules of naming molecular compounds: Replace the ending of the second element with “ide”. While naming molecular compounds prefixes are used to indicate the number of a given element present in the compound like ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.
Solution Summary: The author explains the stock system for naming metal ions with more than one oxidation state in compounds indicated by Roman numeral in parenthesis.
Definition Definition Elements containing partially filled d-subshell in their ground state configuration. Elements in the d-block of the periodic table receive the last or valence electron in the d-orbital. The groups from IIIB to VIIIB and IB to IIB comprise the d-block elements.
Chapter 2, Problem 110RQ
Interpretation Introduction
Interpretation:
The formula of the given compounds are to be written.
Concept Information:
Transition metals have the ability to form more than one positive ion and two ionic compounds with given non-metals.
Stock system is the method for naming the metal ions that have more than one oxidation state in compounds indicated by Roman numeral in parenthesis just after the name of the metal ion.
For naming of ionic compounds containing polyatomic ions, the name of the metal (cation) or a polyatomic cation written first, followed by the name of anion, this can be non-metal, and it this is non-metal (anion), change the suffix of non-metal to -ide, adding at the end of the anion, that is in case of monatomic ions. The hydroxide ion and cynide ion are the exceptions.
Rules of naming molecular compounds:
Replace the ending of the second element with “ide”.
While naming molecular compounds prefixes are used to indicate the number of a given element present in the compound like ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.
Potential Energy (kJ)
1. Consider these three reactions as the elementary steps in the mechanism for a chemical reaction.
AH = -950 kJ
AH = 575 kJ
(i) Cl₂ (g) + Pt (s) 2C1 (g) + Pt (s)
Ea = 1550 kJ
(ii) Cl (g)+ CO (g) + Pt (s) → CICO (g) + Pt (s)
(iii) Cl (g) + CICO (g) → Cl₂CO (g)
Ea = 2240 kJ
Ea = 2350 kJ
AH = -825 kJ
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a. Draw the potential energy diagram for the reaction. Label the data points for clarity.
The potential energy of the reactants is 600 kJ
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0
-200-
-400
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-800-
Reaction Progress
Can u help me figure out the reaction mechanisms for these, idk where to even start
Hi, I need your help with the drawing, please. I have attached the question along with my lab instructions. Please use the reaction from the lab only, as we are not allowed to use outside sources. Thank you!