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Vladimir79 [104]
2 years ago
6

How many SO2 molecules are in 1.76 mol of SO2? How many sulfur atoms and oxygen atoms are there?

Chemistry
1 answer:
Ugo [173]2 years ago
5 0

Answer:

1.76 * 6.02*10^23 = 1.05952*10^24

1.05952*2 = 2.11904 *10^24 oxygen and 1.05952*10^24 sulfur atoms

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What is the difference between dissolving and dissociating and how does this relate to the van't hoff factor?
Nastasia [14]

Solvation is the process of dissolving.

Dissociation is the separation of a substance into charged components during dissolution.

<h3>What is the difference between dissolving and dissociating?</h3>

Dissolution (dissolving), which simply refers to becoming solvated by the solvent, is a subset of dissociation, which is particularly the separation of two charged ions in solution. When the solvent completely envelops each individual solute molecule, solvation takes place. By definition, something that is dissociated in a solvent must be dissolved in it. For instance, the salt dissociates in water, which is how it dissolves in water. In water, ethanol dissolves but does not dissociate; it remains ethanol.

<h3>How does Van't Hoff's factor work?</h3>

The Van't Hoff factor is a tool used to compare the real number of moles of solute injected to create a solution to the apparent amount produced by colloidal characteristics.

<h3>What is the Van't Hoff factor's value?</h3>

Van't Hoff factor's value is:-

  • No association or dissociation                  i=1
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Learn more about Van't Hoff's factor here:-

brainly.com/question/24598605

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6 0
1 year ago
If 23.7 g of Al(OH)3(s) are mixed with 29.5 g of H2SO4(s) and the reaction is run, answer the following questions:
expeople1 [14]

Aluminum hydroxide \text{Al}(\text{OH})_3 can behave as a base and neutralize sulfuric acid \text{H}_2\text{SO}_4 as in the following equation:

2\;\text{Al}(\text{OH})_3 \; (s) + 3\; \text{H}_2\text{SO}_4 \; (aq) \to \text{Al}_2(\text{SO}_4)_3 \; (aq) + 6 \; \text{H}_2\text{O} \; (l) (Balanced)


(a)

n = m/M. Thus the ratio between the number of moles of the two reactants available:

n(\text{Al}(\text{OH})_3, \text{supplied}) / n(\text{H}_2\text{SO}_4, \text{supplied})\\= [m(\text{Al}(\text{OH})_3)/ M(\text{Al}(\text{OH})_3)] / [n(\text{H}_2\text{SO}_4) / M(\text{H}_2\text{SO}_4)]\\= [23.7 / (26.98 + 3 \times(16.00 + 1.008))]/[29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00)]\\\approx 1.01

The value of this ratio required to lead to a complete reaction is derived from coefficients found in the balanced equation:

n(\text{Al}(\text{OH})_3, \text{theoretical}) / n(\text{H}_2\text{SO}_4, \text{theoretical}) = 2/3 \approx 0.667

The ratio for the complete reaction is smaller than that of the reactants available, indicating that the species represented on the numerator, \text{Al}(\text{OH})_3, is in excess while the one on the denominator, \text{H}_2\text{SO}_4, serves as the limiting reagent.


(b)

The quantity of water produced is dependent on the amount of limiting reactants available. 29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00) = 0.301 \; \text{mol} of sulfuric acid is supplied in this reaction as the limiting reagent. 6 moles of water molecules are produced for every 3 moles of sulfuric acid consumed. The reaction would thus give rise to 0.301 \; \text{mol} \times 6/3 = 0.602 \; \text{mol} of water molecules, which have a mass of 0.602 \times (2 \times 1.008 + 16.00) = 10.8 \; \text{g}.


(c)

\text{Percentage Yield}\\= \text{Actual Yield} / \text{Theoretical Yield} \times 100 \; \%\\= 2.21 / 10.8 \times 100 \; \%\\= 20.4 \; \%


(d)

The quantity of \text{Al}(\text{OH})_3, the reactant in excess, is dependent on the number of moles of this species consumed in the reaction and thus the quantity of the limiting reagent available. The consumption of every 3 moles of sulfuric acid, the limiting reagent, removes 2 moles of aluminum hydroxide \text{Al}(\text{OH})_3 from the solution. 0.301 \; \text{mol} of sulfuric acid is initially available as previously stated such that 0.301 \; \text{mol} \times 2/3 = 0.201 \; \text{mol}, or 0.201 \times (26.98 + 3 \time (16.00 + 1.008)) = 15.7 \; \text{g}, of \text{Al}(\text{OH})_3 would be eventually consumed.

23.7 - 15.7 = 8.0 \; \text{g} of \text{Al}(\text{OH})_3 would thus be in excess by the end of the reaction process.

3 0
3 years ago
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