1) Compund Ir (x) O(y)
2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g
3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g
4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g
5) Convert grams to moles
moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles
moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131
6) Find the proportion of moles
Divide by the least of the number of moles, i.e. 0.00656
Ir: 0.00656 / 0.00656 = 1
O: 0.0131 / 0.00656 = 2
=> Empirical formula = Ir O2 (where 2 is the superscript for O)
Answer: Ir O2
Average atomic mass is the weighted average atomic masses with regard to the relative abundance of the isotopes
average atomic mass of Li = relative abundance of Li-6 x mass of Li-6 + relative abundance of Li-7 x mass of Li-7
average atomic mass of Li = (7.42% x 6.0151 a.m.u) + (92.58% x 7.0160 a.m.u)
= 0.446 + 6.495
= 6.941 amu
average atomic mass of Li is 6.941 amu
Using electronegativity difference is a good guide to the ionic/ covalent nature. Large differences indicate greater ionic character, small differences more covalent character. The larger the difference in electronegativity the more ionic properties a bond is said to have. The smaller the difference in electronegativity the more covalent properties a bond is said to have.
Ionic bonding is formed through electrostatic attraction between a cation and anion. Foe example, Sodium fluoride has ionic bonding because it is composed by sodium and Fluorine (a non metal). On the other hand, covalent bonding is characterized by atoms sharing pairs of electrons. For example; methane has covalent bonding; carbon has 4 valence electrons and hydrogen has 1; when they bond they have a total of 8 electrons and satisfies the octet rule.
i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:
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ii. Given the dissolution of some substance
,
the Ksp, or the solubility product constant, of the preceding equation takes the general form
![K_{sp} = [B]^y [C]^z](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BB%5D%5Ey%20%5BC%5D%5Ez)
.
The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.
So, given our dissociation equation in question i., our Ksp expression would be written as:
![K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Cmathrm%7B%5BPb%5E%7B2%2B%7D%5D%20%5BSO_4%5E%7B2-%7D%5D%7D)
.
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iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).
We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:
So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
