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2 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]2 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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Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ

Where h is Planck's constant = 6.626 × 10⁻³⁴ js

c is the speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ

λ = hc / E

λ = (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

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28 days ago

Four students are developing a model to illustrate covalent and ionic substances dissolving in

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Students dealing with ionic bonds comprehend better how to convey what the model should showcase.

Explanation:

Upon dissolving ionic compounds in water, the compounds separate into their constituent ions via a process called dissociation.
The ions become attracted to water molecules, which carry a polar charge.
If the pull between the ions and the water molecules is strong enough to disband the ionic bonds, the compound dissolves.
The ions disperse in the solution, each surrounded by water molecules to inhibit reattachment.
The ionic solution forms an electrolyte, allowing it to conduct electricity.
In contrast, while covalent compounds do dissolve in water, they separate into molecules, not individual atoms.
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This implies that covalent compounds often do not dissolve in water and instead form a distinct layer on top of the water.

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1 month ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

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Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

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2 months ago

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lorasvet [2795]

Answer:

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$87.54 mol O2 x \frac{15}{85}$ = 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2

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How many fluorine atoms are present in 5.85 g of c2f4?

eduard [2782]

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The calculation for the number of molecules per mole is done via the equation,
(0.0585 mols) x (6.022 x 10^23)
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(0.0585 mols) x (6.022 x 10^23)(4)
= 1.41 x 10^23 atoms of F

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1 month ago