Explanation:
(a) The given reaction equation is as follows.
(acidic)
So, here the reduction and oxidation-half reactions will be as follows.
Oxidation-half reaction:
Reduction-half-reaction:
As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.
(acidic)
(b) The given reaction equation is as follows.
(basic)
So, here the reduction and oxidation-half reactions will be as follows.
Reduction-half reaction:
Oxidation-half reaction:
Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.
Therefore, balancing the whole reaction equation in the basic medium as follows.
Answer:
Some properties, such as solubility, melting point, boiling point, and density are independent of the amount of substance being examined. These properties are known as intensive properties and are used to identify a substance. The density of a substance is defined as the mass per unit volume.
Explanation:
<span>Superconductors are materials that are able to conduct electricity especially well because they </span>do not have the resistance of most metals.
Answer:
2. (C) K⁺; 3. (E) Hg⁺; 4. Hg⁺
Explanation:
We must first write the electron configurations of the different species.
(A) Fe²⁺
Fe: [Ar]4s²3d⁶
Fe²⁺: [Ar]3d⁶
When removing electrons from a transition metal ion, you remove the s electrons first.
(B) Cl
Cl: [Ne]3s²3p⁵
(C) K⁺
K: [Ar]4s
K⁺: [Ar]
(D) Cs
Cs: [Xe]6s
(E) Hg⁺
Hg: [Xe]6s²4f¹⁴5d¹⁰
Hg⁺: [Xe]6s4f¹⁴5d¹⁰
2. K⁺ has a noble gas configuration
3. Hg⁺ has electrons in f orbitals.
4. The electron configuration of Au is [Xe]6s4f¹⁴5d¹⁰, not [Xe]6s²4f¹⁴5d⁹, because a filled d subshell is more stable than a filled s subshell.
Thus, Hg⁺ is isoelectronic with Au.
The
chemical reaction is represented as:<span>
2A(g) = B(g) + C(g)
To determine the equilibrium concentration of A, we make use of the equilibrium
constant, Kc, given above. It is expressed as the ratio of the equilibrium
concentrations of the products and the reactants. For this reaction, it is
expressed as:
Kc = [B] [C] / [A]^2
From the problem statement, we are given the following
Kc = 0.035
Volume = 20.0 L
Initial concentrations: [B] = 8.00 mol / 20.0 L = 0.4 M
[C] = 12.00 mol / 20.0 L = 0.6
M
Since the initial reactants are B and C, the reaction is reversed as well as
the Kc.
Kc = [A]^2 / [B][C]
We use the ICE table:
B
C A
I 0.4 0.6
0
C -x -x
+x
------------------------------------------
E 0.4 - x 0.6 - x
x
Kc = x^2 / (0.4-x) (0.6-x) = 0.035
solve for x,
x = 0.07691 = [A]</span>