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1)Prepare n-butanyl bromide from n-butane
2)Prepare Ethyl bromide from n-butane
3)Prepare 1-Butanol from n-butane
4)Prepare 2-Butene from n-butane
1) CH3-CH2-CH2-CH3 + Cl2 ---gt; CH3-CH2-CH2-CH2-Cl + HCl CH3-CH2-CH2-CH2-Cl + KOH (dry) --gt; CH3-CH2-CH2=CH2 CH3-CH2-CH2=CH2 + Br2/H2O2 --gt; CH3-CH2-CH2-CH2-Br CH3-CH2-CH2-CH2-Br + KOH (aq) --gt; CH3-CH2-CH2-CH2-OH + KBr CH3-CH2-CH2-CH2-OH + NaBr/H2SO4...
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The vapor pressure of 1-propanol is 10.0 torr at 14.7 °C. Calculate the vapor pressure at 52.8 °C.Given: Heat of vaporization of 1-propanol = 47.2 kJ/mol
Vapor pressure=34.2atm
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The complete combustion of octane with oxygen, represented by the unbalanced
chemical equation: C8H18 + O2 → CO2 + H2O
10 grams of octane and 10grams of oxygen will yield what mass of water?
Solution:The balanced chemical equation:2C8H18 + 25O2 → 16CO2 + 18H2O According to the chemical equation: n(C8H18)/2 = n(O2)/25 = n(H2O)/18Convert all given information into moles:Moles of octane = n(C8H18) = Mass...
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When 13 g of heptane and 87 g of octane are mixed at 25 degrees Celsius, calculate the vapor pressure for the resulting solution. Vapor pressure of heptane is 31 torr and that of octane is 11 torr.
m(СН3(СН2)5СН3) = 13 g m(C8H8) = 87 g t = 250C = 298K p1 = 31 torr p2 = 11 torr p - ? p = p1 + p2 =...
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For a particular isomer of C8H18, the combustion reaction produces 5099.5 kJ of heat per mole of C8H18(g) consumed, under standard conditions.
C8H18(g)+252O2(g)⟶8CO2(g)+9H2O(g)Δ
The standard enthalpy of formation of this isomer of is -222 kJ/mol
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Calculate the volume of oxygen required to burn 40ml of octane (C8H18) completely. The combustion of octane is ? (2C8H18+25O2 ➡️ 16Co2+18H2O
2C8H18 + 25O2 = 16CO2 + 18H2OOctane's density is 0.703 g cm−340 mL of it weights:40 x 0.703 = 28.12 gM (octane) = 114.23 g/moln = 28.12 / 114.23 =...
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The density of butane, C4H10 , is 2.49×10−3 g/mL . Calculate the number of moles of butane in a 35.9 mL sample.
2.49×10−3 g/mL × 35.9 mL = 0.089391 g Butane molar mass 58.124 g·mol−1 0.089391 / 58.124 g·mol−1 = 0.001538 mol
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How many grams of carbon dioxide (CO2) will be created from the incomplete combustion of 40 g of octane (C8H18)?
2 C8H18 + 25O2→ 16CO2 + 18H2O
M(CO2)=12+2×16=44(g/mol),M(C8H18)=8×12+18×1=114(g/mol).Mass of CO2 may be found from the proportion:x=40×16×44/2×114=123.5(g).
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Butane gas, C4H10, is combusted to produce carbon dioxide and water vapour. If 0.67 mol of butane are reacted, how many moles of water vapour are produced?
2C4 H10 + 13 O2 = 8 CO2 +10 H2On (C4 H10 )= 0,67 mole n (H2O)=10*0.67/2=3,35 moleanswer: 3,35 mole
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Determine the amount of theoretical oxygen and theoretical air needed to burn
a.) 60 kg of butane (C4H10)
b.) 20 kg of propane (C5H12)
c.) 80 kg of octane (C8H18)
A. 2C4H10 + 13O2 ----> 8CO2 + 10H2OMolar mass of C4H10 = (12x4)+10= 58gmol-1molar mass of O2= 16x2=32gmol-12(58)g of C4H10 required 13(32)g of OxygenTherefore 60Kg of C4H10 will require 416x60/116...
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