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Sunny_sXe [5.5K]
2 years ago
13

Consider the ions of potassium (K) and sulfur (S), Write chemical formulas for all possible ionic compounds involving these ions

, using the simplest ratio(s) of potassium (K) and sulfur (S). Keep in mind that the sum of the charges in an ionic compound must equal zero. Answer:​
Chemistry
2 answers:
Nookie1986 [14]2 years ago
6 0

The compounds between the ions K^+ and S^{2-} are K_2S.

<h3>What is an ionic compound?</h3>

An ionic compound is a compound that is composed of ions. Let us note that ionic compounds are always such that the sum of the charges on the ions is zero.

You need two potassium atoms to form an ionic bond with one sulfur atom. Potassium ions are formed by donating and losing their electrons in the outermost shell, the lost electrons are gained by sulphur atom to form 2K^+ and S^{2-}, respectively.

The attraction between the positive ions and negative ions gives rise to the ionic bond and the molecular formula is K_2S

Learn more about ionic compounds:

brainly.com/question/9167977

#SPJ1

attashe74 [19]2 years ago
4 0

The compounds that could exists between the ions K^+ and S^2- are K2S and K2S2.

<h3>What is an ionic compound?</h3>

An ionic compound is a commpound that is composed of ions. Let us note that ionic compounds are always such that the sum of the charges on the ions is zero.

Hence, the compounds that could exists between the ions K^+ and S^2- are K2S and K2S2.

Learn more about ionic compounds:brainly.com/question/9167977

#SPJ1

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Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of
myrzilka [38]

Answer:

167 mL.

Explanation:

We'll begin by calculating the number of moles in 45 g of aluminum (Al). This can be obtained as follow:

Mass of Al = 45 g

Molar mass of Al = 27 g/mol

Mole of Al =?

Mole = mass /Molar mass

Mole of Al = 45/27

Mole of Al = 1.67 moles

Next, the balanced equation for the reaction. This is given below:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

From the balanced equation above,

2 moles of Al reacted with 3 moles of H2SO4.

Next, we shall determine the number of mole of H2SO4 needed to react with 45 g (i.e 1.67 moles) of Al. This can be obtained as:

From the balanced equation above,

2 moles of Al reacted with 3 moles of H2SO4.

Therefore, 1.67 moles of Al will react with = (1.67 × 3)/2 = 2.505 moles of H2SO4.

Thus 2.505 moles of H2SO4 is needed for the reaction.

Next, we shall determine the volume of H2SO4 needed for the reaction. This can be obtained as follow:

Molarity of H2SO4 = 15.0 M

Mole of H2SO4 = 2.505 moles

Volume =?

Molarity = mole /Volume

15 = 2.505 / volume

Cross multiply

15 × volume = 2.505

Divide both side by 15

Volume = 2.505/15

Volume = 0.167 L

Finally, we shall convert 0.167 L to mL. This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.167 L = 0.167 L × 1000 mL / 1 L

0.167 L = 167 mL

Thus, 0.167 L is equivalent to 167 mL.

Therefore, 167 mL H2SO4 is needed for the reaction.

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