Answer:
In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.
Explanation:
In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.
An alkyne contains four carbon atoms.... so if you do 26 multiplied by 4 it equals 104... I do not know if that’s the answer so I apologize if it’s wrong :,)
Answer:
see explanation
Explanation:
The process of ionization to produce cations is endothermic. For formation of Ca⁺² two ionization steps need be illustrated as follows...
1st ionization step: Ca° + 590Kj => Ca⁺ + e⁻
2nd ionization step: Ca⁺ + 1151Kj => Ca⁺² + e⁻
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Net Ionization Rxn: Ca° + 1741Kj => Ca⁺² + 2e⁻
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
Answer:
1.0 M
Explanation:
Reaction equation;
KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)
Concentration of acid CA = ?
Concentration of base CB = 1.0 M
Volume of base VB = 25.60 - 0.50 = 25.1 ml
Volume of acid VB = 25.0 ml
Number of moles of acid NA = 1
Number of moles of base NB =2
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1 * 25.1 * 1/25.0 *1
CA = 1.0 M
