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anygoal [31]
2 years ago
7

Which of the following are considered pure substances?

Chemistry
2 answers:
PolarNik [594]2 years ago
5 0

A. element

and B. compound because they are both considered pure substances

Brut [27]2 years ago
5 0

Answer: Option (A) is the correct answer.

Explanation:

A pure substance is defined a substance that contains atoms of same element and it has constant composition throughout the sample.

For example, sulfur, H_{2}, nitrogen etc are all pure substances.

Whereas a compound is a substance that contains two or more substances which are combined together in a fixed composition.

For example, NaCl is a compound.

When components of solute are uniformly distributed in a solvent then the solution is known as homogeneous solution.

When components of solute are non-uniformly distributed in a solvent then the solution is known as heterogeneous solution.

Thus, we can conclude that elements are considered pure substances.

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lbvjy [14]

If I remember correctly from my science class, it is Protons and Neutrons.

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7 0
3 years ago
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Molecules are likely to dissolve in nonpolar solvents.
masha68 [24]
The answer to this is yes they are likely, true.
4 0
2 years ago
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150
mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after  5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ <em>C</em> NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ <em>C</em> CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ <em>C</em> CH3COOH = 0.0375 M

⇒ <em>C</em> NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ <em>C </em>NaOH = 0.075 M

⇒ <em>C</em> CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ <em>C </em>NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ <em>C </em>NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0
3 years ago
Consider the reaction represented by the equation?2SO2(g) + O2(g) 2SO3(g).?For the system at chemical equilibrium, which of the
rewona [7]

Answer:

b. The amount of SO3(g) decreases and the value for K increases.

Explanation:

Hello,

In this case, for the given reaction:

2SO_2(g) + O_2(g)\rightleftharpoons  2SO_3(g)

The change in the stoichiometric coefficient is:

\Delta \nu=2-2-1=-1

In such a way, since the reagents have more moles than the products, based on Le Chatelier's principle, if the volume is increased, the side with more moles is favored. In addition, since the formation of reagent is favored, K is diminished based on the law of mass action shown below:

K=\frac{[SO_3]_{eq}^2}{[SO_2]_{eq}^2[O_2]_{eq}}

Therefore the answer is:

b. The amount of SO3(g) decreases and the value for K increases.

Best regards.

3 0
3 years ago
NEED HELP ASAP! What is the relationship between changes in air pressure and wind speeds?
saul85 [17]

Answer:

I think its C

Explanation:

3 0
3 years ago
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