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3 years ago

Approximately how many particles are in 3 moles

Chemistry

1 answer:

mash [69]3 years ago

8 0

Answer:

18.066 × 10²³ particles

Explanation:

Given data:

Number of moles = 3 moles

Number of particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 3 moles of substance:

One mole = 6.022 × 10²³ particles

3 mol × 6.022 × 10²³ particles/ 1 mol = 18.066 × 10²³ particles

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

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3 years ago

What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid? what kind of reaction occurs wh

My name is Ann [436]

The kind of reaction that occurs when you mix aqueous solutions of barium sulfide and sulfuric acid is a precipitation reaction.

<h3>Further Explanation</h3>

The chemical reaction between Ba(OH)2(aq) and H2SO4(aq) is given by;

Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(aq) + 2H₂O(l)

This is a type of precipitation reaction, where a precipitate is formed after the reaction, that is Barium sulfate.

<h3>Other types of reaction</h3><h3>Neutralization reactions </h3>

These are reactions that involve reacting acids and bases or alkali to form salt and water as the only products.
For example a reaction between sodium hydroxide and sulfuric acid.

NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)

<h3>Displacement reactions</h3>

These are reactions in which a more reactive atom or ion displaces a less reactive ion from its salt.

Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)

<h3>Redox reactions </h3>

These are reactions that involve both reduction and oxidation occuring simultaneously durin a chemical reaction.
For example,

Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)

Magnesium atom undergoes oxidation while copper ions undergoes reduction.

<h3>Decomposition reactions</h3>

These are type of reactions that involves breakdown of a compound into its constituents elements.

For example decomposition of lead nitrate.

Pb(NO3)2(S) → PbO(s) + O2(g) + NO2(g)

Keywords: Precipitation

<h3>Learn more about: </h3>

Precipitation reaction: brainly.com/question/11194650
Examples of precipitation reactions: brainly.com/question/11194650
Neutralization reactions brainly.com/question/3243813

Level: High school

Subject: Chemistry

Topic: Chemical reactions

Sub-topic: Precipitation reactions

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