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3 years ago

What is the name of the compound Al2(SO3)3?

Chemistry

2 answers:

jek_recluse [69]3 years ago

6 0

Take note that this is S03 and not S04.

S03:Sulfite
S04:Sulfate

So the answer would be Aluminum Sulfite =)

faust18 [17]3 years ago

3 0

Answer:

Name: Aluminum Sulfite

Explanation:

Systematic nomenclature: dialuminium tris [trioxosulfate (IV)]

Oxoanions: anions derived from removing H + from an oxo acid with a vulgar name:

(SO3)3^-2 (sulfite)

Metal cation according to valence number:

Al (III) or Al+3 (as a cation) (Aluminum)

There are different types of salts, in this case we have a simple salt and the nomenclature for binary substances is used when only one class of cation and one class of anion are present

So:

Al2(SO3)3: Aluminum Sulfite

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What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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3 years ago

What is the percent of hydrogen by mass in CH4O

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Answer:

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2 years ago

Find the normality of 0.321 g sodium carbonate in a 250 mL solution.

Radda [10]

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

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= 0.1886eq ÷ 0.2500L

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