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3 years ago

Which of the materials in the chart originally came from fossil fuels?

Chemistry

2 answers:

marin [14]3 years ago

8 0

Nylon the answer is nylon

Nuetrik [128]3 years ago

5 0

Answer:

Polyester

Acrylic

Nylon

Are all made of fossil fuels.

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The balanced equation below represents the decomposition of potassium chlorate.

dezoksy [38]

The oxidation number of chlorine in the reactant can be determined by K ion and O ion. K ion is +1 and O ion is -2. And the Cl is +5. The gas has the greatest entropy and the solid has the least. In the production, there are solid and gas. So it has more entropy than the reactants with solid only.

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3 years ago

What is the total number of moles of oxygen atoms in 1 mole of N2O3

Rama09 [41]

N₂O₃

3 moles oxgyen atoms in 1 mole .

hope this helps!

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2 years ago

How many significant figures are in 340500

vovikov84 [41]

There are 4 significant digits: 3405.

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3 years ago

A swimming pool is filled with 160,000,000 mL of water, how many gallons of water are in the pool. Please show your work, and re

creativ13 [48]

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3 years ago

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl

Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

$\Delta S = S(products)-S(reactants)----(1)$

a) $PCl3(l) + Cl2(g) \rightarrow PCl5(s)$

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) $2HgO(s) \rightarrow 2Hg(l) + O2(g)$

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) $H2(g) \rightarrow 2H(g)$

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) $U(s) + 3F2(g) \rightarrow UF6(s)$

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

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3 years ago