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xxMikexx [17]
3 years ago
11

5. Give one advantage of KIO3 as primary standard.​

Chemistry
1 answer:
alexdok [17]3 years ago
8 0

Answer:

one advantage of KLO3 as a primary standard is that it is used to know concentration of a solution.

Explanation:

The reaction provides confirmation that the solution is at a specific concentration. Primary standards are often used to make standard solutions (a solution with a precisely known concentration

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Which of the following are examples of types of mixtures?
aev [14]
I think it would be homogeneous and heterogeneous
6 0
3 years ago
A cylinder was charged with 1.25 atm of oxygen gas, 6.73 atm of argon, and 0.895 atm of xenon. What is the mole fraction of each
katrin2010 [14]

Considering the Dalton's partial pressure, the mole fraction of each gas is:

  • x_{oxygen}= 0.14
  • x_{argon}= 0.76
  • x_{xenon}= 0.10

<h3>Dalton's partial pressure</h3>

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

P_{T} =P_{1} +P_{2} +...+P_{n}

where n is the amount of gases in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

P_{A} =x_{A} P_{T}

In summary, the total pressure in a mixture of gases is equal to the sum of partial pressures of each gas.

Mole fraction of each gas

In this case, you know that:

  • P_{oxygen }= 1.25 atm
  • P_{argon}= 6.73 atm
  • P_{xenon}= 0.895 atm
  • P_{T} =P_{oxygen} +P_{argon}+P_{xenon}= 1.25 atm + 6.73 atm + 0.895 atm= 8.875 atm

Then:

  • P_{oxygen} =x_{oxygen} P_{T}
  • P_{argon} =x_{argon} P_{T}
  • P_{xenon} =x_{xenon} P_{T}

Substituting the corresponding values:

  • 1.25 atm= x_{oxygen} 8.875 atm
  • 6.73 atm= x_{argon} 8.875 atm
  • 0.895 atm= x_{xenon} 8.875 atm

Solving:

  • x_{oxygen}= 1.25 atm÷ 8.875 atm= 0.14
  • x_{argon}= 6.73 atm÷ 8.875 atm= 0.76
  • x_{xenon}= 0.895 atm÷ 8.875 atm=0.10

In summary, the mole fraction of each gas is:

  • x_{oxygen}= 0.14
  • x_{argon}= 0.76
  • x_{xenon}= 0.10

Learn more about Dalton's partial pressure:

brainly.com/question/14239096

brainly.com/question/25181467

brainly.com/question/14119417

#SPJ1

3 0
2 years ago
The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
meriva

Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

8 0
3 years ago
What kind of questions can science answer
IrinaK [193]

Answer:

hypothesis

Explanation:

a question that can be answered wity a hypothesis is a question a scientist can answer

5 0
3 years ago
I need help on dis chem shi if y’all can answer fa me I would appreciate dat
elena55 [62]
I think it’s A I’m not 100% sure but I mean it’s worth a try
3 0
3 years ago
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