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VikaD [51]
3 years ago
5

Use the IUPAC nomenclature rules to give the name for this compound - Al2S3.

Chemistry
2 answers:
jonny [76]3 years ago
6 0
Aluminium Sulfide
According to rules the positive specie is named first and the negative specie is named last.
Inessa [10]3 years ago
6 0

Answer : The IUPAC name of Al_2S_3 is aluminum sulfide.

Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Ionic compound : It is a type of compound that is made up of the ions of the two different elements in which one element is a metal and another element is a non-metal.

The nomenclature of ionic compounds is given by:

  • Positive is written first.
  • The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

The given compound is, Al_2S_3

Here aluminum is having an oxidation state of (+3) called as Al^{3+} cation and sulfur S^{2-} is an anion with oxidation state of (-2). Thus, they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Al_2S_3 called as aluminum sulfide.

Hence, the IUPAC name of Al_2S_3 is aluminum sulfide.

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nexus9112 [7]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

To calculate the amount of heat absorbed or released, we use the following equation:

q=mc\Delta T    .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of  water = 4.186 J/g ° C      

\Delta T = Change in temperature

  • <u>For Trial 1:</u>

We are given:

m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J

Putting values in equation 1, we get:

q=30g\times 4.186J/g^oC\times 40^oC

q = 5023.2 J

  • <u>For trial 2:</u>

We are given:

m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J

Putting values in equation 1, we get:

q=40g\times 4.186J/g^oC\times 10^oC

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = (5023.2-1674.4)J=3347J

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

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c₁=5.28M
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v₁-?

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Answer:

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