Answer:
f. Sn^4+
c. second
e. Al^3+
d. third
Explanation:
This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.
Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.
However, there will be occurrence of precipitation after the 1st step1.
So, the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.
Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
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Answer:
In first shell only 2 electrons are present in hydrogen only one electron is present in valence shell. To complete its duplet hydrogen can share lose or gain only one electron to form chemical bond.
Hydrogen can form only one bond.
Explanation:
Answer:
B
The increase in the atomic number
