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Svetach [21]
4 years ago
12

Why do real gases not behave exactly like ideal gases?

Chemistry
1 answer:
mihalych1998 [28]4 years ago
8 0

Answer:

Real gas particles have significant volume

Real gas particles have more complex interactions than ideal gas particles.

Explanation:

An ideal gas is an imaginary concept and a gas behaves almost ideally at certain pressure and temperature conditions.

The gas in real deviates from the ideal behavior as some of the assumptions made for ideal gases are not true in case of real gases.

Real gas particles have significant volume as compared to vessel unlike ideal gases.

There are interactions present in between real gas molecules at high pressure conditions.

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In the laboratory you dissolve 12.3 g of chromium(ii) sulfate in a volumetric flask and add water to a total volume of 375. ml.
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Hope it cleared your doubt.

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3 years ago
How is the oxidation state of a transition metal determined from the chemical formula ?
Vera_Pavlovna [14]

Answer:

Explanation:

In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.

We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.

The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.

To ascertain the oxidation state, we have to comply with some rules:

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For example, let us find the oxidation state of Cr in Cr₂O₇²⁻

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We see that the oxidation number of Cr, a transition metal in the given ion is +6.

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3 years ago
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Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

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[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

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3 years ago
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