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3 years ago

Explain the relationship between the roman numeral and the charge used in naming ionic compounds.

1 answer:

3 0

The Roman numerals are used to show the oxidation number of some transition metals because some elements have more than one possible oxidation state. So basically it's there to indicate the charge of the element.

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TRUE or FALSE. adding a solute like sodium chloride to water will always increase the freezing point

Afina-wow [57]

I think that is false but don't quote me

8 0

3 years ago

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Which one of the following substances would be the most soluble in CCl4?

maria [59]

Answer: Option (d) is the correct answer.

Explanation:

Carbon tetrachloride $(CCl_{4})$ is a non-polar solvent. Whereas out of the given options, $Na_{2}SO_{4}$ , $H_{2}O$ , $CH_{3}CH_{2}CH_{2}CH_{2}OH$ , and HI are all polar molecules.

On the other hand, only $C_{4}H_{10}$ is non-polar molecule.

Also it is known that like dissolves like.

So, being non-polar $CCl_{4}$ will dissolve the give alkane, $C_{4}H_{10}$ .

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3 years ago

What is the symbol for Lithium, Iron, and Helium

jonny [76]

Lithium=li

Iron=Fe

Helium=He

I don't say u must have to mark my ans as brainliest but if u think it has really helped u plz don't forget to thank me....

8 0

3 years ago

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The mass of the atom is found by adding is protons and neutrons the electrical charge of an atom is found by adding the charge i

Hitman42 [59]

When you combine protons and neutrons in the electrical charge they combine to make electricity

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3 years ago

In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area

storchak [24]

First we must write a balanced chemical equation for this reaction

$Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)$

The mole ratio for the reaction between $HNO_3$ and $Na_2CO_3$ is 1:2. This means 1 moles of $Na_2CO_3$ will neutralize 2 moles $HNO_3$ . Now we find the moles of each reactant based on the mass and molar mass.

$2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3$

$2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3$

$\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}$

The $Na_2CO_3$ was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.

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3 years ago