A flood that affects the environment where natural rubber is produced would severely hinder rubber production. In order to greatly limit production, a flood would need to destroy a significant portion of rubber trees. Natural rubber is crucial for manufacturing synthetic polymers. If the rubber supply is compromised (due to the disruption of its ecosystem caused by a flood), there would be a substantial decline in the availability of synthetic polymers.
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The response to your inquiry is: option B. 0.25 atm
Explanation:
To solve this issue, the combined gas law must be applied:
P₁V₁ = P₂V₂ / T₁T₂
The data is as follows: P1 = 0.99 atm, V1 = 2 L, T1 = 273 K, P2 =?, V2 = 4 L, T2 = 137 K.
By isolating P2 in the equation, you find
P2 = P1V1T2 / T1V2. Substituting in the numbers gives: P2 = (2 x 0.99 x 137)/(273 x 4). The resulting P2 equates to approximately 0.25 atm.
Answer: The process of heating a crucible to eliminate moisture from a hydrate.
Explanation:
The available choices are:
a. Heating a solvent to aid in the dissolution of a solute.
b. Heating a solid in isolation to remove moisture.
c. Bringing water to a boil for use in a water bath.
d. Heating a crucible to eliminate moisture from a hydrate.
Possible actions that can be done on a hot plate include:
a. Heating a solvent to assist a solute in dissolving.
b. Heating a solid in isolation to dry it.
c. Heating water to boiling for a water bath.
However, it's important to note that using a hot plate for heating a crucible to remove water from a hydrate is not advisable. Silica or ceramic materials are not meant to be heated on a hot plate.
Consequently, the correct procedure is heating a crucible to remove water from a hydrate.
A total of 1.505×10^23 lead atoms
In the lungs, the volume of lead equals the total lung volume, which is 5.60L
1 mole corresponds to 22.4L
Thus, 5.6L of lead converts to 5.6/22.4 = 0.25 moles
According to Avogadro's law
1 mole of lead contains 6.02×10^23 lead atoms
Thus, 0.25 moles of lead equates to 0.25×6.02×10^23 = 1.505×10^23 lead atoms
3 first significant figure
6 second significant figure
5 third significant figure
4 cannot exceed 5, so retain 5 instead of increasing it to 6
0.0365
