Answer:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
Explanation:
To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;
Be = 2, 2
F = 2, 7
Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.
When Be loses two electrons it becomes isoelectronic with He;
Be → Be²⁺ + 2e⁻
Also, when fluorine gains an electron, it becomes isoelectronic with Ne;
F + e⁻ → F⁻
This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.
Hence;
Be²⁺ + 2F⁻ → BeF₂
Answer:
specific heat = 0.951 j/g·°C
Explanation:
Heat flow equation => q = m·c·ΔT
q = heat flow = 4817 joules
m = mass in grams = 140 grams Aluminum
c = specific heat = ?
ΔT = Temperature Change in °C = 98.4°C - 62.2°C = 36.2°C
q = m·c·ΔT => c = q/m·ΔT = 4817j/(140g)(36.2°C) = 0.951 j/g·°C
Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
Answer:
Explanation:
Hello,
In this case, since the reaction between potassium hydroxide and nitric acid is:
We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:
That in terms of molarities and volumes is:
Thus, solving the molarity of the base (KOH), we obtain:
Regards.
