How would a flood be a limitation for synthetic polymers that rely on natural rubber in its production?

2 answers:

4 0

Answer : Synthetic polymer production is directly not dependent on the climatic effects; but in the floods if the rubber plants are harmed or the production is hampered or by limited availability of rubber plants due to floods can deeply effect the synthetic polymer manufacturing industry which are dependent on natural rubber plants. So, indirectly it may hamper the production of synthetic polymers.

Gnom [1K]3 years ago

3 0

Answer: just put synthetic polymer

Explanation: Synthetic polymer production is directly not dependent on the climatic effects; but in the floods if the rubber plants are harmed or the production is hampered or by limited availability of rubber plants due to floods can deeply effect the synthetic polymer manufacturing industry which are dependent on natural rubber plants. So, indirectly it may hamper the production of synthetic polymers.

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If a buffer solution is 0.130 M in a weak acid (K_a = 1.7 x 10^-5) and 0.590 M in its conjugate base, what is the pH?

Serggg [28]

Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43

8 0

3 years ago

How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of

xeze [42]

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

$moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}$

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three: