14. Two or more substances that are together but do not combine to form a new, pure substance.

1 answer:

6 0

The answer would be a mixture. A mixture is when two or more substances are combined to create heterogeneous mix, where the substances do not combine with each other in an equal or even state, but just sit together.

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Match the term with its description. match term definition acid

jok3333 [9.3K]

B and d will work out and a and c will also work out

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3 years ago

What is the net ionic equation for 2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 6NaOH

tatuchka [14]

Answer:

2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3

Explanation:

First of all, let us balance the equation to give;

2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 3NaOH

Now, we can observe the presence of positive Sodium ions (Na+) and negative hydroxyl ions (OH-) on both left and right sides of the equation.

Now, the two ions will cancel out. These ions are not really involved in the overall reaction and thus do not require being written in the overall equation. Hence, the overall net ionic reaction can now be written as:

2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3

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Arrange the elements in order of increasing atomic radius. Use the periodic table to help you.

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Answer:

F, S, Mg, Ba

Explanation:

Fluorine has a small atomic radius. These go up until Ba, which has a larger radius

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3 years ago

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Oxidation of methane

Mandarinka [93]

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➪Methane (CH4) is oxidized with molecu- lar oxygen (O2) to carbon dioxide (CO2).

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A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m

kogti [31]

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that the gas is O2 gas

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Under the same conditions;

the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$