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Alja [10]
3 years ago
15

List the metals that form only one type of ion (that is, metals whose charge is invariant from one compound to another). What ar

e the group numbers of these metals
Chemistry
1 answer:
NeTakaya3 years ago
3 0

this question is in reference to the formation and naming of ionic compounds. Specifically, they want you to give examples of three ionic compounds with a metal to nonmetal ratio of 2 to 1. That means we need to have two metal atoms to metal ions, which are typically cat ions for every one non metal atom or an ion. In order for this to occur, we need to have the metal with half the charge of the nonmetal or the non metal with double the charge of the metal. So an example might be something like sodium sulfide. Sodium has one valence electron. It can give up sulfur needs to valence electrons in order to achieve an octet. So we need to. Sodium seems to give up one electron each to total so that sulfide can achieve an octet. Another one might be potassium oxide. Similar scenario. We've got potassium giving up one valence electron oxygen requiring too. So we need to potassium to supply the to valence electrons that oxygen needs to achieve an octet and lithium. Also in Group one A and alkali metal wants to give up just one valence electron to achieve an octet well to achieve, I guess a duet to be more like helium, and so it gives up one. If we have two of them, then we can provide the to valence electrons that sulfur needs. So this is sodium sulfide, potassium oxide and lithium sulfide. Remember when we name Ionic compounds? We named the Cat Ion with the name of the element and the anti on with the name of the Element, but with the ending oven of ID, a suffix of ID because each one of the cat ions donated their valence electrons to the anti on so the an ion could achieve an octet. Then all of the's will have an octet of valence electrons. Sulfur had six sodium had one. There were two of them, so we have a total of eight.

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3 0
3 years ago
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3 years ago
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8 0
3 years ago
The thermite reaction reacts iron (III) oxide, Fe2O3, with aluminium powder,Al, the form aluminium oxide, Al2O3 and iron, Fe.
lara [203]

Answer:

<u><em>This answer assumes that the strated "16.0g of iron" was meant to be 16.0 grams of iron(III) oxide.</em></u>

Explanation:

To start, the thermite equation must be balanced.

I find:

1Fe2O3 + 2Al = 1Al2O3 + 2Fe

This tells us we need 2 moles of Al for every 1 mole of Fe2O3.  

Now calculate the moles of each reactant:

Moles Fe2O3:  16.0 g/159.7 g/mole = <u>0.100 moles Fe2O3</u>

Moles Al:  8.1 /26.98 g/mole = <u>0.300 moles Al</u>

The balanced equation says that in order to react all of the Fe2O3 we'd need twice that amount (in moles) of the Al.  (0.100 moles Fe2O3)*(2) = 0.200 moles Al.

<u>Which of the two reactants is the limiting reagent?</u>

We have more than enough moles of Al to react with 0.10 moles of Fe2O3.  (We have 0.300 moles Al and all we need is 0.200 moles to react with the 0.10 moles of Fe2O3.  <em>Fe2O3 is the limiting reagent.</em>

<u><em>Calculate the maximum mass of iron of iron that could be formed using these quantities of reactants.</em></u>

The balanced equation tells us that we will obtain 2 moles of Fe for every 1 mole of Fe2O3 consumed.  Since Fe2O3 is the limiting reagent, we will assume that it completely reacts.  That means 0.1 moles of Fe2O3 is reacted.  Since we expect twice that many moles of Fe, we should obtain 0.200 moles of Fe.  At 55.85 g/mole, we should obtain:

(0.200 moles Fe)*(55.85 g Fe/mole Fe) = 11.2 grams Fe

<em></em>

<em></em>

5 0
2 years ago
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