The combinations of Strontium and chlorine that are possible are only those in which strontium and chlorine combine in the ratio of 1:2.
1) The possible combinations are
- 4 strontium atoms and 8 chlorine
strontium atoms and
chlorine atoms
2) The combinations that are not possible are
- 20 strontium atoms and 60 chlorine atoms
- 130 billion strontium atoms and 195 billion chlorine
We have to work out the ratio of Strontium and chlorine in each of the given combinations in the question. Only the combinations in which the ratio of Strontium and chlorine is 1:2 is possible.
First case:
4 strontium atoms and 8 chlorine atoms gives a Strontium and chlorine ratio of 1:2 so it is possible.
Second case:
20 strontium atoms and 60 chlorine atoms gives a Strontium and chlorine ratio of 1:3 hence it is not possible.
Third case:
strontium atoms and chlorine atoms gives a Strontium and chlorine ratio of 1:2 hence it is possible.
Fourth case:
130 billion strontium atoms and 195 billion chlorine atoms gives a Strontium and chlorine ratio of 1:1.5 hence it is not possible.
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3rd one sorry if it’s wrong
I know water will change the shape to match the container its in if that is your question
Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
I believe that the answer is ionic
