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Colt1911 [192]
3 years ago
6

The bond angles marked a, b, and c in the molecule below are about __________, __________, and __________, respectively. Select

correct answer. Select correct answer. 120°, 109.5°, 120° 90°, 90°, 90° 109.5°, 120, 109.5° 109.5°, 109.5°, 109.5° 109.5°, 109.5°, 90°
Chemistry
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

109.5°, 120°, 109.5°

Explanation:

The bond angles that are marked as a, b and c in the given molecules is in the order of :

Angle a : 109.5°

Nitrogen is $sp^3$ hybridized and the geometry is tetrahedral. So the bond angle is 109.5°

Angle b : 120°

Carbon is $sp^2$ hybridized, so the shape is trigonal planer and the bond angle is 120°

Angle c : 109.5°

Carbon is $sp^3$ hybridized and the geometry is tetrahedral. The bond angle is 109.5°

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A sample of natural rubber (200.0 g) is vulcanized, with the complete consumption of 4.8 g of sulfur. Natural rubber is a polyme
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Answer: 1.3% many crosslinks as isoprene units,  

Explanation:

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mass pf natural rubber= 200.0g

mass of sulphur = 4.8g

molar mass of sulphur =32g/mol

molar mass of isoprene = C5H8=( 12x5) +(1x8)= 68g/mol

Solution: we first find no of moles present in each  using

no of moles = \frac{mass}{molarmass}

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to find % crosslinked units, we have  

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