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bonufazy [111]
3 years ago
10

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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The standard emf for the cell using the overall cell reaction below is +2.20 V:
ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: 2Al-6e^{-}\rightarrow 2Al^{3+}

Reduction: 3I_{2}+6e^{-}\rightarrow 6I^{-}

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Overall: 2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}

Nernst equation for this cell reaction at 25^{0}\textrm{C}-

E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}

where n is number of electrons exchanged during cell reaction, E_{cell}^{0} is standard cell emf , E_{cell} is cell emf , [Al^{3+}] is concentration of Al^{3+} and [Cl^{-}] is concentration of Cl^{-}

Plug in all the given values in the above equation -

E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V

So, E_{cell}=2.32V

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2 years ago
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I am Lyosha [343]

Answer:

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7 0
3 years ago
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BlackZzzverrR [31]
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A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called
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Explanation:
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