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3 years ago

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

Chemistry

2 answers:

snow_tiger [21]3 years ago

6 0

Answer:

Sample Response: A flood would affect the availability of a material used in making a synthetic polymer. If a flood destroys rubber trees, the availability of natural rubber would decrease, which would in turn decrease the production of synthetic polymers.

Explanation:

Right on edge

Zepler [3.9K]3 years ago

3 0

A flood, if it hits the environment of the natural rubbers, would destroy how the rubber is being produced. to have a large amount of limitation, the flood would destroy a large percentage of rubber trees. This natural rubber is needed to make synthetic polymers. Without the rubber (because of damages to it's ecosystem through the flood), there would be a limited supply, and a substancial drop on synthetic polymers.

hope this helps

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Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according

Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

6 0

3 years ago

Consider the first-order reaction described by the equation At a certain temperature, the rate constant for this reaction is 5.8

zubka84 [21]

<u>Answer:</u> The half life of the reaction is 1190.7 seconds

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

k = rate constant of the reaction = $5.82\times 10^{-4}s^{-1}$

$t_{1/2}$ = half life of the reaction = ?

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{5.82\times 10^{-4}s^{-1}}=1190.7s$

Hence, the half life of the reaction is 1190.7 seconds

8 0

3 years ago

Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0

3 years ago

Why is biodiversity important?

VARVARA [1.3K]

A or the first answer

3 0

3 years ago

Read 2 more answers

An aqueous solution has a volume of 2.0 L and contains 36.0g of glucose. If the Molar Mass of glucose is 180g/mol, what is the m

choli [55]

Answer:

moles = 36/180 = 0.2 moles

molarity = 0.2/2 = 0.1 mol/dm3

3 0

2 years ago