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Elza [17]
2 years ago
13

What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

Chemistry
1 answer:
Morgarella [4.7K]2 years ago
5 0

This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5%​ as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

\% H=\frac{m_H*1}{M_{HF}}*100\%

Where m_H stands for the atomic mass of hydrogen and M_{HF} for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%

Learn more about percent compositions: brainly.com/question/12247957

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Which is an example of a non aqueous solution
damaskus [11]

Answer: gasoline

Explanation:

An aqueous solution is one which contains water as a solvent. A non- aqueous solution is the one in which the solvent of the solution is a liquid but it is not water.

Among the options given, gasoline is the correct option. This is because of the fact that gasoline is a crude oil which does not include water as an ingredient.

3 0
3 years ago
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Discuss elements and compounds by completing the following paragraph
pychu [463]

Elements are substances that are made up of the same atoms which are capable of taking part in a chemical reaction.

There are different types of elements which are represented by symbols gotten from the first letter or the first and any other letter in the name of the element.

Examples of elements include:

  • Hydrogen (H)
  • Carbon (C)
  • Nitrogen (N)
  • Sodium (Na)

When two or more of these elements combine together through a chemical bond, it leads to the formation of compounds.

Example of a compound includes:

  • NaCl: The element sodium combine, through electrochemical bonding, with another element chlorine to form the compound sodium chloride.

Learn more here:

brainly.com/question/17571315

3 0
3 years ago
4 points
Ilya [14]

The vertical columns are groups

The horizontal rows are periods

4 0
3 years ago
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A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2
Nezavi [6.7K]

Answer:

  • The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

<u>1) Chemical reaction.</u>

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

  • sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

  • NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

  • 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

<u>2) Mole ratio</u>

  • 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

<u>3) Starting amount of oxalic acid</u>

  • mass = 28 mg = 0.028 g
  • molar mass = 90.03 g/mol
  • Convert mass in grams to number of moles, n:

        n = mass in grams / molar mass = 0.028 g / 90.03 g/mol =  0.000311 mol

<u>4) Titration</u>

  • Volume of base: 28.4 mL = 0.0248 liter
  • Concentration of base: x (unknwon)

  • Number of moles of acid: 2.52 mol (calculated above)
  • Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622

That means that there are 0.000622 moles of NaOH (solute)

<u>5) Molarity of NaOH solution</u>

  • M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.

5 0
3 years ago
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Solid cadmium sulfide reacts with an aqueous solution of sulfuric acid . Express your answer as a balanced chemical equation. Id
max2010maxim [7]

Explanation:

When solid cadmium sulfide reacts with an aqueous solution of sulfuric acid then the reaction will be as follows.

          CdS(s) + H_{2}SO_{4}(aq) \rightarrow CdSO_{4}(aq) + H_{2}S(g)

Hence, ionic equation for this reaction is as follows.

      CdS(s) + 2H^{+}(aq) + SO^{2-}_{4}(aq) \rightarrow Cd^{2+}(aq) + SO^{2-}_{4}(aq) + H_{2}S(g)

Therefore, net ionic equation for this reaction is as follows.

      CdS(s) + 2H^{+}(aq) \rightarrow Cd^{2+}(aq) + H_{2}S(g)

8 0
3 years ago
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