Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.
Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.
=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 
The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
Ammonium carbonate will form 3 moles of ions.
Methyl alcohol will form 0 moles of ions.
Methane will form 0 moles of ions.
Aluminum sulfite will form 3 moles of ions.
Hydrobromic acid will form 2 moles of ions.
Explanation:
One mole of ammonium carbonate will form 3 moles of ions when dissolved in water.
(NH₄)₂CO₃ (s) + H₂O (l) → 2 NH₄⁺ (aq) + CO₃²⁻ (aq) + H₂O (l)
One mole of methyl alcohol will form 0 moles of ions when dissolved in water.
(actually it form ions because of its acidic behavior but they are at the order of 10⁻⁷ moles, however in the framework of this question we may say that there are none)
One mole of methane will form 0 moles of ions when dissolved in water.
Methane does not react with water (in normal conditions) so will not form ions.
One mole of aluminum sulfite will form 3 moles of ions when dissolved in water.
Al₂SO₃ (s) + H₂O (l) → 2 Al₃⁺ (aq) + SO₃²⁻ (aq) + H₂O (l)
One mole of hydrobromic acid will form 2 moles of ions when dissolved in water.
HBr (l) + H₂O (l) → Br⁻ (aq) + H₃O⁺ (aq)
Learn more about:
solvation of ions
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To make our solution more systematic, let's convert all units that is consistent with the units of R which is 0.0821 L-atm/mol-K.
For pressure: 760 torr = 1 atm
388 torr * 1 atm/760 torr = 0.5105 atm
For volume, 1 μL = 10⁻⁶ L
0.206 μL * 10⁻⁶ L/1 μL = 2.06×10⁻⁷ L
For temperature,
T = 45 + 273 = 318 K
For mass, 1 ng = 10⁻⁹ g
206 ng * 10⁻⁹ g/1 ng = 2.06×10⁻⁷ g
Assuming ideal gas,
PV=nRT
(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)
n = 4×10⁻⁹ mol
Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol
<em>Molar mass = 51.14 g/mol</em>
The value of Δ H butane (g) = -124.7 kJ/mol
The value of Δ H CO2 (g) = -393.5 kJ/mol
The value of Δ H H2O (g) = -241.8 kJ/mol
Mass of butane, m = 8.30 gm
Molar mass of butane is 58 gm/mol
Consider the reaction,
C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O
Calculating the value of Δ H° rxn:
ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)
Substituting the values we get,
Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)
= -1574 -1209 + 124.7
= -2783 - 124.7
= -2658.3 kJ/mol
Now, calculate the number of moles of butane in 8.30 gm.
Number of moles = mass/molar mass
= 8.30 / 58
= 0.143 moles
Thus, the total energy released in the reaction is,
Q = number of moles × ΔH° rxn
= 0.143 × (2658.3)
= 380.14 kJ
Hence, the total heat released in the reaction is 380.14 kJ.
