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3 years ago

Classify unopened root beer as an element, a compound, a homogeneous mixture, or a heterogeneous mixture

Chemistry

1 answer:

maxonik [38]3 years ago

4 0

The different state of material based on its ingredients are-

1. Compound: The compound is in which there are more than one molecule in which chemical bond is there.

2. Mixture: It is basically presence of two or more compounds in which there is no chemical bonds.

The mixture may be of two types (i) homogeneous and (ii) Heterogeneous. In the homogeneous mixture there is uniformity of the mixed compound (like air in water) and some chemical process is needed to separate the compounds. Whereas heterogeneous mixtures (like chalk in water) are easily separable by any physical process.

In the root beer which is a soft drink which is obtained in different flavor sometime with caffeine and without caffeine. The unopened bottle of root beer is homogeneous mixture.

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frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

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