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sweet-ann [11.9K]
3 years ago
11

What is the henry's law constant for co2 at 20∘c? express your answer to three significant figures and include the appropriate u

nits?
Chemistry
2 answers:
bulgar [2K]3 years ago
8 0

The value of Henry’s law constant for {\text{C}}{{\text{O}}_{\text{2}}} at 20{\text{ }}^\circ {\text{C}} is \boxed{3.70 \times {{10}^{ - 2}}{\text{ L}} \cdot {\text{atm}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}}.

Further explanation:

Solubility

It is that property of substance by virtue of which it becomes able to dissolve in other substances. It is measured in terms of the maximum amount of solute that can be dissolved in the given amount of solvent.

Henry’s Law

According to this law, solubility of gas dissolved in the liquid is directly related to the partial pressure of gas. High partial pressure means high solubility and vice-versa.

Mathematically,

{{\text{S}}_{{\text{gas}}}} \propto {{\text{P}}_{{\text{gas}}}}  …… (1)                                                                                    

To remove the proportionality constant in equation (1), constant known as Henry’s constant is incorporated and equation (1) modifies to,

{{\text{S}}_{{\text{gas}}}} = {{\text{k}}_{\text{H}}}{\mathbf{ \times }}\;{{\text{P}}_{{\text{gas}}}}         …… (2)                                                                      

Here,

{{\text{S}}_{{\text{gas}}}} is the solubility of gas.

{{\text{k}}_{\text{H}}} is Henry’s constant.

{{\text{P}}_{{\text{gas}}}} is the pressure of gas.

Equation (2) can be rearranged in order to calculate Henry’s constant \left( {{{\text{k}}_{\text{H}}}} \right) and equation (2) becomes,

{{\text{k}}_{\text{H}}} = \dfrac{{{{\text{S}}_{{\text{gas}}}}}}{{{{\text{P}}_{{\text{gas}}}}}}\;    …… (3)                                                                                  

At 20{\text{ }}^\circ {\text{C}} , the value of Henry’s law constant for {\text{C}}{{\text{O}}_{\text{2}}} up to three significant figures is 3.70 \times {10^{ - 2}}{\text{ L}} \cdot {\text{atm}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}.

Learn more:

  1. What is the concentration of alcohol in terms of molarity? brainly.com/question/9013318
  2. Determine whether solubility will increase, decrease or remains the same: brainly.com/question/2802008

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: solubility, gas, Henry’s law, partial pressure, solubility, dissolve, CO2, three significant figures, 3.70*10^-2 L atm/mol, high partial pressure, high solubility.

Gennadij [26K]3 years ago
3 0
Actually, Henry's Law is an empirical value. It means that it was not obtained out of raw calculations or correlations. This was gathered from experimental results. Hence, you can search its data. At standard temperature of 25°C (298 K),

k = k°e^[2400(1/T - 1/T°)], where k° = 29.4 L·atm/mol
Substituting the values so that T would be in 20°C or 293 K,
k = (29.4 L·atm/mol)e^[2400(1/293 - 1/298)]
k = 33.7 L·atm/mol
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An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
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Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

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b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

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Identify the oxidation state of Ba 2 + . +2 Identify the oxidation state of S in SO 2 . +2 Identify the oxidation state of S in
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Answer:

The oxidation state of Ba in cation Ba²⁺ is +2

The oxidation state of S in SO₂, is +4

The oxidation state of S in anion sulfate (SO₄⁻²) is +6

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Explanation:

We define oxidation state as the number which can be negative or positive that  

indicates the number of electrons that the atom has accepted or transferred.

All the elements in ground state has 0 as oxidation state.

This numbers are very important for redox reaction which are balanced by the ion electron method.

When the elements gain electrons, the element is being reduced so the oxidation state decreases.

When the elements release electrons, the element is oxidized so the oxidation state increases.

We have to think, that global charge of a compound is 0, for example in the ZnSO₄.

The sulfate anion has a global charge of -2 because it has released 2 protons, it came from the sulfuric acid (H₂SO₄). As the global charge is -2, oxygen acts with -2, and the anion has 4 atoms so the global charge of O is -8. Definetly S, has +6 as oxidation state.

In the SO₂, oxygen acts with -2 and there are 2 atoms in the compound, so the global charge is 0 and the global charge for O  is -4. Therefore S must act with +4.

Ba²⁺ is an element of group 2 and has a tendency to form a cation, so it can release electrons for that purpose.  At least, it can release 2 e⁻, that's why the oxidation state is +2. It can complete the octet rule and it will be isoelectronic with Xe.

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Answer:

C. Its oxidation number increases.

Explanation:

  • <em><u>Oxidation is defined as the loss of electrons by an atom while reduction is the gain of electrons by an atom</u></em>.
  • Atoms of elements have an oxidation number of Zero in their elemental state.
  • When an atom looses electrons it undergoes oxidation and its oxidation number increases.
  • For example, <em><u>an atom of sodium (Na) at its elemental state has an oxidation number of 0. When the sodium atom looses an electrons it becomes a cation, Na+, with an oxidation number of +1 , the loss of electron shows an increase in oxidation number from 0 to +1.</u></em>
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