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AleksandrR [38]
3 years ago
11

Bubble wrap is made of a series of air pockets sealed in a sheet of plastic and is often used to protect fragile objects during

shipping. Bubble wrap tends to pop when stepped onto or squeezed. How can you explain this using Boyle's law?
Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

Explanation:

Boyle's law states that at constant temperature:

P \alpha \frac{1}{V}

PV = constant

Where: P is the pressure of a given mass of gas, and V is its volume.

The bubble wrap has a confined volume of air at constant temperature. Since a gas is easily compressible, the volume of the gas in the wrap decreases with increasing pressure. This implies that there is an inverse proportion relation between the pressure on and volume of the air.

The bubble wrap pops when stepped onto or squeezed because the pressure is high and the volume can not be compressed further. Thus, the elastic limit of the seal is exceeded. This leads to an escape of the confined air.

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²⁻(aq)

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(aq) + 2I⁻

(aq)

<u>Explanation:</u>

S₂O₈²⁻

(aq) + 2I⁻

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To measure the rate of this reaction we must measure the rate of concentration change of one of  the reactants or products. To do this, we will include (to the reacting S₂O₈

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i) a small amount of sodium thiosulfate, Na₂S₂O₃,

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The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂  as soon as it is formed.

2S₂O₃²⁻

(aq) + I₂(aq) → S₄O₆²⁻

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(aq)

This reaction is much faster than the previous, so the conversion of I2 back to I⁻  is  essentially instantaneous.

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Answer:

Option B. +3 and +6

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Answer:

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