Answer:
Four different reagent where used 1) BH3 & 2) H2O2/NaOH to produce butanol, followed by 3)PBr3 to produce Bromobutane and finally 4) NaOBu to produce the dibutylether
Explanation:
Please find attached the step by step reaction
Answer:
0.172 M
Explanation:
The reaction for the first titration is:
First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume</em>:
- 19.6 mL * 0.189 M = 3.704 mmol HCl
As one HCl mol reacts with one NaOH mol, <em>there are 3.704 NaOH mmoles in 25.0 mL of solution</em>. With that in mind we <u>determine the NaOH solution concentration</u>:
- 3.704 mmol / 25.0 mL = 0.148 M
As for the second titration:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
We <u>determine how many NaOH moles reacted</u>:
- 34.9 mL * 0.148 M = 5.165 mmol NaOH
Then we <u>convert NaOH moles into H₃PO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 5.165 mmol NaOH * = 1.722 mmol H₃PO₄
Finally we <u>determine the H₃PO₄ solution concentration</u>:
- 1.722 mmol / 10.0 mL = 0.172 M
The elements fluorine (F), chlorine (CL),and Iodine are all parts of the same _group___ on the periodic table
Treat compound b with a strong base to get compounds C & D!
Answer:
The heat needed to convert 1 kg of feed water at 20°C into dry saturated steam at a pressure of 9 bar is 2690.19 kJ/kg.
Explanation:
Step 1 : Obtain the enthalpy of staurated steam and enthalpy of evaporation at 9 bar pressure from the steam table:
From steam table, at 9 bar, the enthalpy of saturated water = 742.83 kJ /kg
enthalpy of evaporation = 2031.1 kJ /kg
Step 2: Calculate the enthalpy of dry saturated steam:
Enthalpy of dry saturated steam = enthalpy of saturated water + enthalpy of evaporation
= 742.83 + 2031.1 = 2773. 93 kJ/ kg
Step 3: Calculate the enthalpy of 1 kg of feed water at 20°C
Enthalpy of 1 kg of feed water = c * ( T2 -T1)
= 4.187 * (20-0)
= 83.74 kJ /kg
Step 4 : calculate the heat needed by 1 kg of feed water at 20°C to be converted to dry saturated steam at 9 bar:
Heat needed = Enthalpy of dry saturated steam - enthalpy of feed water
Heat needed = 2773.93 kJ/kg - 83.74 kJ/kg
Heat = 2690.19 kJ/kg