In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
Here is what Golgi looks like, so that you can look at the picture and describe it.
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Glad I could help, and good luck!
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C. a plane for forming a scientific hypotheses
<u>Given:</u>
Volume of Na2CO3 = 250 ml = 0.250 L
Molarity of Na2CO3 = 6.0 M
Volume of CaF2 = 750 ml = 0.750 L
Molarity of CaF2 = 1.0 M
<u>To determine:</u>
The mass of CaCO3 produced
<u>Explanation:</u>
Na2CO3 + CaF2 → CaCO3 + 2NaF
Based on the reaction stoichiometry:
1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3
Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles
Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles
CaF2 is the limiting reagent
Thus, # moles of CaCO3 produced = 0.750 moles
Molar mass of CaCO3 = 100 g/mol
Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g
Ans: Mass of CaCO3 produced = 75 g
