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Chapter 3: Problem 29

Which formulas are not correct? For each that is not correct, write the correct formula. (a) Ammonium phosphate; \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PO}_{4}\) (b) Barium carbonate; \(\mathrm{Ba}_{2} \mathrm{CO}_{3}\) (c) Aluminum sulfide; \(\mathrm{Al}_{2} \mathrm{S}_{3}\) (d) Magnesium sulfide; MgS

Short Answer

Expert verified
(a) and (b) are incorrect. Correct: Ammonium phosphate: \((\text{NH}_4)_3\text{PO}_4\), Barium carbonate: \(\text{BaCO}_3\).

Step by step solution

01

Verify Formula for Ammonium Phosphate

Ammonium has the formula \( ext{NH}_4^+\) and phosphate has the formula \( ext{PO}_4^{3-}\). To balance the charges, we need three ammonium ions to match one phosphate ion: \[\left(\mathrm{NH}_4\right)_{3}\mathrm{PO}_{4}\]The provided formula is \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PO}_{4}\), which is incorrect.
02

Verify Formula for Barium Carbonate

Barium has the formula \(\text{Ba}^{2+}\) and carbonate has the formula \(\text{CO}_3^{2-}\). Equal charges mean that they balance one to one: \[\mathrm{BaCO}_{3}\]The provided formula is \(\mathrm{Ba}_{2} \mathrm{CO}_{3}\), which is incorrect.
03

Verify Formula for Aluminum Sulfide

Aluminum has the formula \(\text{Al}^{3+}\) and sulfide has the formula \(\text{S}^{2-}\). To balance charges, two aluminum ions are needed for three sulfide ions:\[\mathrm{Al}_{2}\mathrm{S}_{3}\]This matches the provided formula, so it is correct.
04

Verify Formula for Magnesium Sulfide

Magnesium has the formula \(\text{Mg}^{2+}\) and sulfide has the formula \(\text{S}^{2-}\). They balance one to one:\[\mathrm{MgS}\]This matches the provided formula, so it is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Ammonium Phosphate
Ammonium phosphate is a compound containing ammonium ions and phosphate ions. The chemical formula for ammonium is \(\text{NH}_4^+\), a positively charged ion. Phosphate, on the other hand, is \(\text{PO}_4^{3-}\), a negatively charged ion.
To create a neutral compound, the positive and negative charges must balance. Since phosphate has a 3- charge, three ammonium ions (each with a 1+ charge) are needed to balance the charge of one phosphate ion. Thus, the correct formula for ammonium phosphate is \(\left(\text{NH}_4\right)_3\text{PO}_4\).
Keep in mind these key points:
  • The charge of the ions determines how many of each ion are needed.
  • Balancing charges ensures the chemical formula is neutral.
  • Ammonium phosphate involves a coordination of three ammonium and one phosphate ion.
Decoding Barium Carbonate
Barium carbonate combines barium ions and carbonate ions. Barium has the formula \(\text{Ba}^{2+}\), which is a positive ion. The carbonate ion is \(\text{CO}_3^{2-}\), a negative ion with a 2- charge.
When formulating barium carbonate, these ions must be in equal amounts to neutralize the charges, resulting in the formula \(\text{BaCO}_3\).
The formula should have:
  • One barium ion for one carbonate ion.
  • A neutral compound from balanced charges (2+ and 2-).
  • Understanding the charges helps in determining the correct formula quickly.
Aluminum Sulfide Formula Explained
Aluminum sulfide comprises aluminum and sulfide ions. Aluminum ions bear a 3+ charge, represented as \(\text{Al}^{3+}\), whereas sulfide ions have a 2- charge, written as \(\text{S}^{2-}\).
To achieve a neutral compound, the charges need balancing through a combination of two aluminum ions and three sulfide ions as per the formula \(\text{Al}_2\text{S}_3\).
Key balance points are:
  • Two aluminum ions provide a total positive charge of 6+.
  • Three sulfide ions counterbalance with a total negative charge of 6-.
  • Ensuring the total positive and negative charges equal helps affirm the formula's accuracy.
Exploring Magnesium Sulfide
Magnesium sulfide involves magnesium and sulfide ions. Magnesium, featuring a \(\text{Mg}^{2+}\) ion, pairs with sulfide, \(\text{S}^{2-}\), both carrying a 2 charge but of opposite signs. These ions naturally balance one-to-one because their charges offset completely.
The formula \(\text{MgS}\) shows an equal balance of positive and negative charges, forming a neutral compound.
Points of interest include:
  • One magnesium ion pairs with one sulfide ion.
  • Achieving neutrality with matching charge magnitudes is crucial.
  • This straightforward pairing simplifies understanding ionic compounds.

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Most popular questions from this chapter

Write the formula for the compound formed from the following pairs of ions: (a) Iron(III) ion and hydroxide ion (b) Barium ion and chloride ion (c) Calcium ion and phosphate ion (d) Sodium ion and permanganate ion

Show how each chemical change obeys the octet rule. (a) Lithium forms Li \(^{+}\) (b) Oxygen forms \(\mathrm{O}^{2-}\)

Sodium borohydride, \(\mathrm{NaBH}_{4},\) has found wide use as a reducing agent in organic chemistry. It is an ionic compound composed of one sodium ion, \(\mathrm{Na}^{+},\) and one borohydride ion, \(\mathrm{BH}_{4}^{-}\) (a) How many valence electrons are present in the borohydride ion? (b) Draw a Lewis structure for the borohydride ion. (c) Predict the \(\mathrm{H}-\mathrm{B}-\mathrm{H}\) bond angles in the borohydride ion.

Table 3.2 shows the following ions of copper: \(\mathrm{Cu}^{+}\) and \(\mathrm{Cu}^{2+} .\) Do these violate the octet rule? Explain.

Answer true or false. (a) An ionic bond is formed by the combination of positive and negative ions. (b) An ionic bond between two atoms forms by the transfer of one or more valence electrons from the atom of higher electronegativity to the atom of lower electronegativity. (c) As a rough guideline, we say that an ionic bond will form if the difference in electronegativity between two atoms is approximately 1.9 or greater. (d) In forming NaCl from sodium and chlorine atoms, one electron is transferred from the valence shell of sodium to the valence shell of chlorine. (e) The formula of sodium sulfide is \(\mathrm{Na}_{2} \mathrm{S}\) (f) The formula of calcium hydroxide is CaOH. (g) The formula of aluminum sulfide is AlS. (h) The formula of iron(III) oxide is \(\mathrm{Fe}_{3} \mathrm{O}_{2}\) (i) Barium ion is \(\mathrm{Ba}^{2+}\), and oxide ion is \(\mathrm{O}^{2-}\); therefore, the formula of barium oxide is \(\mathrm{Ba}_{2} \mathrm{O}_{2}\)
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