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schepotkina [342]
2 years ago
5

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

Chemistry
2 answers:
snow_tiger [21]2 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]2 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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2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

5 0
2 years ago
Based on the crystal-field strengths cl- < f- < h2o < nh3 < h2nc2h4nh2, which octahedral ti (iii) complex below has
kati45 [8]
<span>Based on the crystal field strength, Cl ligand would give the longest d-d transition when complexed with Ti(III). as this is the weak field ligand and would cause minimum splitting of d orbitals.</span>
8 0
2 years ago
Show that the Newton has the units of mass times acceleration
Dimas [21]

F = ma = (kg)(m/s2) = kg ´ m/s2 N

hope this helps :D

7 0
2 years ago
An isotope undergoes radioactive decay by emitting radiation that has no mass. What other characteristic does the radiation have
Advocard [28]

Answer : Option D) No charge

Explanation : An isotope undergoes radioactive decay by emitting radiation that has no mass. The radiation will not have any charge as it does not has any mass it will not emit a radiative charge.

It is known that there are some unstable radioactive isotopes which has no mass and the radiation thus has no charge in it.

7 0
2 years ago
Read 2 more answers
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
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