All categories

4 years ago

How many additional electrons can fit into the n=3 shell of sulfur

Chemistry

1 answer:

Sever21 [200]4 years ago

8 0

2. <span>(Note that sulfur is 2 columns from the right-hand end of the periodic table)</span>

You might be interested in

A *Blank* reaction occurs when one compound reacts and is broken down into different elements or simpler compounds

In-s [12.5K]

A dissociation reaction occurs.lets say pcl5=pcl3+cl2

4 0

3 years ago

Read 2 more answers

Write the balanced nuclear equation.

Vladimir [108]

Answer:

pag pinag balik tad moyan tama magiging tama sagot nya promise jslsh

3 0

3 years ago

An isotope contains 11 protons, 10 electrons, and 12 neutrons. What is the identity of the isotope?

Nataly [62]

Neon-22

Depending on the type of notation the question wants but that is generic term for it.

8 0

4 years ago

Read 2 more answers

A negatively charged particle has too many ___

oee [108]

Answer:

electrons

Explanation:

8 0

4 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.