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4 months ago

A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound

Chemistry

1 answer:

lorasvet [2.7K]4 months ago

5 1

Answer:

The empirical formula is: C₂H₃O

Explanation:

The empirical formula, often referred to as the “minimum formula,” provides the simplest way to represent a chemical compound, illustrating both the present elements and their minimum integer ratios in terms of atoms.

The mass percentage composition indicates the proportion by mass of each element in a compound.

Using 100 g of the compound as a reference, the percentage can be expressed in grams. Thus, presuming 100 g of the compound consists of 53.46 g of C, 6.98 g of H, and 39.56 g of O.

Considering the molecular masses of each component, one can derive the number of relative atoms for each element:

C: $53.46 g *\frac{1 mol}{12.01 g } = 4.45 moles$

H: $6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles$

O: $39.56 g *\frac{1 mol}{16g } = 2.47 moles$

Subsequently, divide each value by the smallest one:

C: $\frac{4.45 moles}{2.47 moles}= 1.80$

H: $\frac{6.91 moles}{2.47 moles}= 2.8$

O: $\frac{2.47 moles}{2.47 moles}=1$

Approximating decimals to the nearest integer gives:

C: 2

H: 3

O: 1

Therefore the empirical formula is: C₂H₃O

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castortr0y [3046]

Answer:

Explanation:

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r=1.2

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$A=A_0e^{\lambda t}$

where,

$A_0$ is the initial activity of the substance

Now, solve for t

$-\lambda t=In\frac{A}{A_0}$

$t=-\frac{1}{\lambda} In(\frac{A}{A_0} )$

$=-\frac{1}{\lambda} In(\frac{A}{\lambda r(\frac{m_c}{m_a} )} )$

since,

$A_0=\lambda r(\frac{m_c}{m_a} )$

$=-\frac{1}{\lambda} In(\frac{A\ m_a}{\lambda r m_c} )$

Thus, the age of the artifact is

$=-\frac{1}{\lambda} In(\frac{A\ m_a}{\lambda r m_c} )$

$=-\frac{1}{1.21\times 10^{-4}} In(\frac{(9.25)(2.32\times 10^{-26}}{1.21\times 10^{-4}(\frac{1}{3.15569\times10^7} )(1.2\times 10^{-12})(0.100)}} )\\\\=6303.4 \ years$

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3 months ago

In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What

Anarel [2989]

Answer:

The force is 38503.5N.

Explanation:

From the problem, we determine:

P (pressure) = 5.00 atm.

Next, to find the force in Newtons (N), we must convert 5 atm into N/m², as shown:

1 atm equals 101325 N/m².

So, 5 atm equals 5 x 101325 = 506625 N/m².

A (the piston area) = 0.0760 m².

Pressure signifies force per unit area, mathematically represented as

P = F/A.

From this, we find F = P × A.

F = 506625 × 0.0760.

Therefore, F = 38503.5N.

Thus, the piston experiences a force of 38503.5N.

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4 months ago

The symbol for xenon (Xe) would be a part of the noble gas notation for the element antimony. cesium. radium. uranium.

castortr0y [3046]

Noble gas notation serves as a condensed form of indicating electron configurations. This notation employs the symbol for the preceding noble gas in the electron configuration of an element. For antimony, the noble gas prior is Kr, which means Xe is not used in its electron configuration. Similarly, for radium, the prior noble gas is Rn, whereas, for uranium, it is also Rn. However, for cesium, the preceding noble gas is Xe, thus it is utilized in the noble gas notation for Sb, specifically written as: Cs: [Xe] 6s.

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3 months ago

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Enter the net ionic equation representing solid chromium (iii) hydroxide reacting with nitrous acid. express your answer as a ba

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The net ionic equation for the reaction between chromium (III) hydroxide and nitrous acid is:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

Additional Details

An electrolyte dissociates into ions in solution.

Chemical equations can also be represented with ionic species.

Strong electrolytes (fully ionized) are written as separate ions, whereas weak electrolytes (partially ionized) remain as intact molecules.

In ionic equations, spectator ions are those unchanged by the chemical process—they are present both before and after the reaction.

Removing these spectators results in the net ionic equation.

Gases, solids, and water (H₂O) are written as molecules, without ionization.

Therefore, only dissolved compounds are represented by their ions (aq).

The problem involves chromium (III) hydroxide reacting with nitrous acid.

The reaction occurring is:

Cr(OH)₃ (s) + 3HNO₂ (aq) ⇒ Cr(NO₂)₃ (aq) + 3H₂O (l)

Chromium (III) hydroxide is a solid and remains un-ionized, as does water.

Thus, the ionic equation is:

Cr(OH)₃ (s) + 3H⁺ (aq) + 3NO₂⁻ (aq) ⇒ Cr³⁺ (aq) + 3NO₂⁻ (aq) + 3H₂O (l)

The ion 3NO₂⁻ is a spectator ion; removing it yields the net ionic equation:

Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)

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