Chemistry Some Basic Concepts of Chemistry

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3 months ago

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V
Vishal Baghel

Contributor-Level 10

Mass of C15H30 = volume * Density

= 1000 m l  * 0.756 gm/ m l  

C15H30 + 22.5 O2   15CO2 + 15H2O

No. of moles of C15H30 = 7 5 6 2 1 0 moles

No. of moles of O2 required =   ( 2 2 . 5 * 7 5 6 2 1 0 ) m o l e s

Mass of O2 required = 22.5 *   7 5 6 2 1 0 * 3 2 = 2 5 9 2 g m

No. of moles of CO2 liberated = 15 *   ( 7 5 6 2 1 0 ) moles

Mass of O2 liberated =   1 5 * 7 5 6 2 1 0 * 4 4 = 2 3 7 6 g m

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Using : PV = nRT

1.5 * 416 = n * 0.083 * 300

n = 25mol =  W M

25 =  1 0 0 M

So, molar mass, M = 4 g/mol.

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3 months ago

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Vishal Baghel

Contributor-Level 10

V =  n R T P = 0 . 9 0 * 0 . 0 8 2 1 * 3 0 0 1 8 * 3 2 * 7 6 0 = 2 9 . 2 1 2 9

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alok kumar singh

Contributor-Level 10

Empirical formula is C5H7N

Empirical mass = 81

Molecular mass = 162

So, molecular formula is C10H14N2

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3 months ago

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Payal Gupta

Contributor-Level 10

 CH3OH (l)+32O2 (g)CO2 (g)+2H2O (l)

Δn=0.5

ΔH=ΔE+ΔnRT=7260.5*8.31000*300=7261.24=727.24727kJ/mol

Hence, x = 727 (the nearest integer)

New answer posted

3 months ago

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alok kumar singh

Contributor-Level 10

Number of Fe atoms = 0.34100*3.356*6.022*1023

= 1.206 * 1020

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