All categories

3 years ago

What are the rules for naming covalent compounds

1 answer:

6 0

1.Remove the ending of the second element, and add "ide" just like in ionic compounds
2.When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound
3.If there is only one of the first element, you can drop the prefix
4.If there are two vowels in a row that sound the same once the prefix is added (they "conflict"), the extra vowel on the end of the prefix is removed

You might be interested in

Calculate the molarity of 65.3 grams of sugar(342.3 g/mol), c12h22011 in 750 ml of solution.

k0ka [10]

Answer:

0.254 M

Explanation:

If the formula mass of sugar is 342.3 g/mol, and there are 65.3 g, then there are 65.3/342.3=0.190768 mol.

Also, 750 mL = 0.750 L.

molarity = (moles of solute)/(liters of solution)

molarity = 0.190768/0.750 = 0.254 M

8 0

1 year ago

Nonrenewable energy sources come out of the ground as liquids, gases,

elixir [45]

Answer:

TRUE

Explanation:

Production of Hydrocarbons from Natural Gas is as stated below:

Natural gas liquids include propane, butane, pentane, hexane, and heptane, but not methane and not always ethane, (may include it sometimes.) since these hydrocarbons need refrigeration to be liquefied.

4 0

2 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.