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Minchanka [31]
2 years ago
12

Hello people ~

Chemistry
2 answers:
Klio2033 [76]2 years ago
8 0
<h3>CHEMISTRY</h3>

What is the most basic aromatic amine’s common name?

a) Benzenamine

b) Benzylamine

c) Aniline

d) Aminobenzene

#BRAINLYEVERYDAY

Pavel [41]2 years ago
8 0

Answer:

<em><u>BENZYLAMINE </u></em>

Explanation:

  • <em><u>most basic aromatic amine’s common </u></em><em><u>name </u></em><em><u>is </u></em><em><u>BENZYLAMINE</u></em>
  • <em><u>C6H5C2</u></em><em><u>NH2</u></em><em><u> </u></em><em><u>IS </u></em><em><u>FORMULA </u></em>

<em><u>O</u></em><em><u>PTION </u></em><em><u>B </u></em><em><u>~</u></em>

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A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
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Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

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