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cestrela7 [59]
3 years ago
8

What is most likely oxidation state of aluminum (AI)?

Chemistry
1 answer:
blagie [28]3 years ago
7 0
Aluminum has three oxidation states. The most common one is +3. The other two are +1 and +2. One +3 oxidation state for Aluminum can be found in the compound aluminum oxide, Al2O3.
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A hydrocarbon, containing only C and H, can react with excess oxygen gas to produce carbon dioxide and water. The result of the
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I believe the answer you are looking for would be HC

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NEED HELP ASAP!!!!! PLEASE HELP!!!!! NEED HELP ASAP!!!!!
r-ruslan [8.4K]

Answer:

CH4 - Methane

B2Si - Diboron monosilicide

N2O5 - Dinitrogen pentoxide

CO2 - Carbon dioxide

Explanation:

When it comes to naming covalent compounds, there are several rules.

The name is derived based on the formula. For example, N2O5. The first element is nitrogen. To the name of the element, you add the prefix that tells us how many of its atoms are in the compound. In this case, there are two atoms, which means that the prefix will be <em>di</em>- (dinitrogen). The second element is oxygen. You are supposed to take only the root of the second element's name and then add the prefix denoting the number of its atoms and the suffix <em>-ide</em> (pentoxide). This is how we'll get dinitrogen pentoxide.

The only exception is methane (CH4), which is an organic compound. Organic compounds are named using the IUPAC nomenclature.

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3 years ago
What is the percent ionization of a monoprotic weak acid solution that is 0.188 M? The acid-dissociation (or ionization) constan
djyliett [7]

Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is 3.59\times 10^{-4}\%

Explanation:

Dissociation of weak acid is represented as:

HA\rightleftharpoons H^+A^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.188 M and \alpha = ?

K_a=2.43\times 10^{-12}

Putting in the values we get:

2.43\times 10^{-12}=\frac{(0.188\times \alpha)^2}{(0.188-0.188\times \alpha)}

(\alpha)=3.59\times 10^{-6}=3.59\times 10^{-4}\%

Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is 3.59\times 10^{-4}\%

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