Answer:
fluorine (f) lies in group 17 and 2 period
Answer:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Electron_Configuration
Explanation:
Express the oxidation state numerically (e.g., +1).
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Ions
In an ion, the sum of the oxidation states is equal to the overall ionic charge. Note that the sign of the oxidation states and the number of atoms associated with each oxidation state must be considered. In OH−, for example, the oxygen atom has an oxidation state of −2 and the hydrogen atom has an oxidation state of +1, for a total of (−2)+(+1)=−1.
Part C
What is the oxidation state of an individual phosphorus atom in PO33−?
Express the oxidation state numerically (e.g., +1).
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Part D
What is the oxidation state of each individual carbon atom in C2O42−?
Express the oxidation state numerically (e.g., +1).
a. AlCl₃ ⇒ limiting reactant(smaller ratio)
Cu ⇒ excess reactant
b. the mass of leftover reactant : 7.207 g
<h3>Further explanation</h3>
Given
25 g Cu
25 g AlCl3
Required
a. the excess and limiting reactants
b. the mass of leftover reactant
Solution
Reaction
3Cu + 2AlCl₃ ⇒ 3CuCl₂ + 2Al
mol Cu(Ar = 63.5 g/mol) :
mol = mass : Mw
mol = 25 : 63.5
mol = 0.394
mol AlCl3(MW=133,34 g/mol) :
mol = 25 : 133,34 g/mol
mol = 0.187
mol ratio to reaction coefficient Cu : AlCl₃ =
AlCl₃ ⇒ limiting reactant(smaller ratio)
Cu ⇒ excess reactant
b. the mass of leftover reactant :
mol Cu = 3/2 x 0.187 = 0.2805
mol left = 0.394 - 0.2805 = 0.1135
mass = 0.1135 x 63.5 = 7.207 g
Answer:
234 KJ ≡ 55.887 Kcal
Explanation:
∴ J ≡ Kg/m².s² ≡ N.m = 0.2389 cal
∴ cal = 4.187 J
⇒ 234 KJ * ( 1000 J / KJ ) * ( cal / 4.187 J ) * ( Kcal / 1000 cal ) = 55.887 Kcal