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Crank
3 years ago
10

Match the description to the type of mixture.

Chemistry
1 answer:
Softa [21]3 years ago
7 0

Answer:

1-homo

2-hetero

3-hetero

4-homo

Explanation:

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H2O (s) ⇒ H2O What would be the best argument against writing this in the form of a chemical equation?
Goshia [24]

Answer: Since this is merely a Physical Change, it is improper to write it as a chemical equation.

5 0
3 years ago
What does Hess's law state about the total enthalpy change for a reaction?OA. It depends on the reaction pathway.B. It can only
Dimas [21]
<h2>Answer:</h2>

Option D is correct. It is independent of the reaction pathway.

<h2>Explanations:</h2><h2>What is Hess's law?</h2>

Hesslaw states that the enthalpy change of a reaction does not change regardless whether the reaction takes place in a single or multiple reaction pathways.

This shows that the total entalpy change of a reaction does not depend on the reaction pathway.

8 0
1 year ago
Which among the following statements is FALSE? Group of answer choices None of the above A liquid has a definite volume; but it
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Answer: None of the above statements is false.

Explanation:

In a solid substance, particles are closely held together due to which a solid substance has definite shape and volume. Therefore, solids are also incompressible in nature.

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In gases, the particles are held by Vander waal forces due to which they move rapidly from one place to another. Hence, gases are highly compressible in nature.

Thus, we can conclude that none of the given statements are false.

4 0
3 years ago
5 Which two elements are combined in magnesium oxide?
Vitek1552 [10]

Answer:

magnesium and oxygen

Explanation:

6 0
3 years ago
Fatty acids spread spontaneously on water to form a monomolecular film. A solution containing 0.10 mm^3 of a fatty acid is dropp
lara31 [8.8K]

Divide the volume by the area. Using scientific makes things a bit cleaner.

0.10\,\mathrm{mm}^3 = 10^{-1}\,\mathrm{mm}^3

400.\,\mathrm{cm}^2 = 4\times10^2\,\mathrm{cm}^2

Then

\dfrac{10^{-1} \,\mathrm{mm}^3}{4\times10^2\,\mathrm{cm}^2} \cdot \dfrac{\left(\frac{1\,\rm m}{10^3\,\rm mm}\right)^3}{\left(\frac{1\,\rm m}{10^2\,\rm cm}\right)^2} = \dfrac{10^{-1}\times10^{-9} \,\mathrm m^3}{4\times10^2\times10^{-4}\,\mathrm m^2} = \dfrac{10^{-10}}{4\times10^{-2}}\,\mathrm m \\\\ ~~~~~~~~= 0.25\times10^{-8}\,\mathrm m

Now, 1 m = 10⁹ nm, so

0.25 \times10^{-8}\,\mathrm m \cdot \dfrac{10^9\,\rm nm}{1\,\rm m} = 0.25\times10^1\,\mathrm{nm} = \boxed{2.5\,\rm nm}

8 0
1 year ago
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