All categories

3 years ago

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.

You might be interested in

Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0

nexus9112 [7]

Answer: The correct answer is Option b.

Explanation:

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

For Trial 1:

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

For trial 2:

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.

4 0

3 years ago

This is a chemistry question that i don't know how to do. Need some help

koban [17]

C₀=8.10M
c₁=5.28M
v₀=1.58 L
v₁-?

n=c₀v₀=c₁v₁

v₁=c₀v₀/c₁

v₁=8.10*1.58/5.28=2.42 L

5 0

3 years ago

During the experiment a student precipitated and digested the BaSO4. After allowing the precipitate to settle, they added a few

OlgaM077 [116]

Answer:

Incomplete precipitation of barium sulfate

Explanation:

The student has precipitated and digested the barium sulfate on his/her side. But on the addition of $BaCl_2$ in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When $BaCl_2$ is added, there are still sulfate ions present in the solution with combines with $BaCl_2$ and forms $BaSO_4$ and the formation of this precipitate makes the solution cloudy.