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Andru [333]
2 years ago
6

Which law states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constan

t?
Boyle’s law
Dalton’s law
Charles’s law
Avogadro’s law
Chemistry
1 answer:
slava [35]2 years ago
7 0

The law that states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constant is Avogadro’s law. That is option D.

<h3>What is Avogadro’s law?</h3>

Avogadro’s law states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constant.

Mathematically the Avogadro’s law can be represented by the following:

V1/n1= V2/n1

Where,

V1 = first volume

n1 = first amount of gas

V2 = second volume

n2 = second amount of gas

Learn more about gas laws here:

brainly.com/question/25587265

#SPJ1

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Answer:

A volumetric pipette is used to measure 25.0cm3 of 2.0mol/dm3 aqueous sodium hydroxide into aconical flask.

A burette is filled with dilute sulfuric acid.

Write a balanced symbol equation for the reaction between these chemicals when we neutralise.

Explanation:

The reactants of the given reaction are:

sodium hydroxide and dilute sulfuric acid.

The balanced chemical equation of the reaction is:

H_{2}SO_{4} (aq)  + 2NaOH(aq) -> Na_{2} SO_{4} (aq)+ 2H_{2} O(l)

Acid and base raect with each other and form salt and water.

This reaction is called a neutralization reaction.

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3 years ago
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4 0
3 years ago
The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(aq) What
ryzh [129]

Answer : The mass of PbI_2 precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of PbI_2 = 461.01 g/mole

First we have to calculate the moles of NaI.

\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles

Now we have to calculate the moles of PbI_2.

The balanced chemical reaction is,

Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)

From the balanced reaction we conclude that

As, 2 moles of NaI react to give 1 mole of PbI_2

So, 0.042 moles of NaI react to give \frac{0.042}{2}=0.021 moles of PbI_2

Now we have to calculate the mass of PbI_2.

\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2

\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g

Therefore, the mass of PbI_2 precipitate produced will be, 9.681 grams.

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