Chapter 4, Problem 102GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Cloth can be waterproofed by coating it with a silicone layer. This is done by exposing the cloth to (CH3)2SiCl2 vapor. The silicon compound reacts with OH groups on the doth to form a waterproofing film (density = 1.0 g/cm3) of [(CH3)2SiO]n, where n is a large integer number.n (CH3)2SiCl2 + 2n OH– → 2n Cl– + n H2O + [(CH3)2SiO]nThe coating is added layer by layer, with each layer of [(CH3)2SiO]n being 0.60 nm thick. Suppose you want to waterproof a piece of doth that is 3.00 square meters, and you want 250 layers of waterproofing compound on the cloth. What mass of (CH3)2SiCl2 do you need?

Interpretation Introduction

Interpretation:

The mass of (CH3)2SiCl2 needed to waterproof the given quantity of material has to be calculated.

Concept introduction:

• The mass of any material can be determined from its volume and density.  The mathematical relationship between mass, volume and density of material is,

Mass=Density×volume

• For chemical reaction balanced chemical equation is written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
Explanation

Cloth can be waterproofed by coating it with a silicone layer and it is done by exposing the cloth to (CH3)2SiCl2 vapor.

The silicon compound reacts with OH groups on the cloth to form a waterproofing film of [(CH3)2SiO]n where n is a large integer number.

The balanced equation for this reaction is,

â€‚Â n(CH3)2SiCl2â€‰+â€‰2nâ€‰â€‰OH-â€‰â€‰â†’â€‰â€‰2nCl-â€‰+â€‰nH2Oâ€‰+[(CH3)2SiO]n

The coating is added layer by layer, with each layer of [(CH3)2SiO]n being 0.60â€‰â€‰nm thick.

For 250â€‰â€‰ coats total thickness will be 0.6â€‰â€‰nmâ€‰Ã—â€‰â€‰250â€‰â€‰=â€‰â€‰â€‰150â€‰â€‰nmâ€‰â€‰thick

â€ƒâ€‚Â Â â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰1â€‰nmâ€‰=â€‰1â€‰Ã—10-7â€‰m150â€‰â€‰nmâ€‰â€‰

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