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3 years ago

How does adding a non-volatile solute to a pure solvent affect the boiling point of the pure solvent?

1 answer:

7 0

The correct answer is the second statement. The solvent will have a higher boiling point. Adding a non-volatile solute to a pure solvent will increase the boiling point of the solvent. This solution exhibit colligative properties. Colligative properties depend on the amount of solute dissolved in a solvent. These set of properties do not depend on the type of species present.

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Who is credited with creating the first periodic table

sergeinik [125]

Dmitri Mendeleev is the answer

5 0

3 years ago

What type of physical change is it when a liquid turns into a solid and a solid turns into a liquid

goblinko [34]

Answer:

Freezing and Melting (Fusion)

Explanation:

4 0

3 years ago

Choose the words to finish the sentence. After learning about the law of conservation of mass, Sammy became interested in balanc

cestrela7 [59]

The chemical equation is unbalanced and synthesized.

<h3></h3><h3>What is a chemical equation?</h3>

A chemical equation is described as the symbolic representation of a chemical reaction in the form of symbols and chemical formulas.

In a chemical equation, the reactant entities are given on the left-hand side and the product entities is shown on the right-hand side with a plus sign between the entities in both the reactants and the products, and an arrow that indicates towards the products to show the direction of the reaction.

We can conclude that in the chemical equation shown is unbalanced because both amounts of the individual elements and compounds do not reflect on the reactant and product side.

Learn more about chemical equations at: brainly.com/question/11231920

#SPJ1

The complete question is below:

After learning about the law of conservation of mass, Sammy became interested in balancing equations. He knew that the symbol for aluminum was Al and silver tarnish was Ag2S. He also knew that mixing the two chemicals yielded pure silver, or Ag, in an aluminum sulfide solution. Here is the equation showing this reaction:

3 Ag2S + 2 Al → 6 Ag + Al2S3

This equation is (synthesis / unbalanced / replacement / balanced), and it represents a(n) (unbalanced / balanced / synthesized / replaced) chemical reaction.

answer choices:

synthesis; balanced

balanced; replacement

unbalanced; synthesized

balanced; balanced

6 0

1 year ago

A certain hydrocarbon has a molecular formula of C9H16. Which of the following is not a structural possibility for this hydrocar

snow_tiger [21]

Answer: Option (d) is the correct answer.

Explanation:

It is given that molecular formula is $C_{9}H_{16}$ . Now, we will calculate the degree of unsaturation as follows.

Degree of unsaturation = $C_{n} - \frac{\text{monovalent}}{2} + \frac{\text{trivalent}}{2} + 1$

= $9 - \frac{16}{2} + 1$

= 9 - 8 + 1

= 2

As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.

Yes, this compound could be an alkyne as for alkyne D.B.E = 2.

But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.

Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.

6 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago