N = 4 to n = 3 is the right answer, so it' none of the above
Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.
<h3>How can iron (iii) chloride be formed from iron fillings?</h3>
Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:
- Iron fillings + Cl₂ → FeCl₃
Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.
In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.
Learn more about iron (iii) chloride at: brainly.com/question/14653649
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<h3>
Answer:</h3>
0.35 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial volume as 35.0 mL or 0.035 L
- Initial molarity as 12.0 M
- Final volume is 1.20 L
We are required to determine the final molarity of the solution;
- Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
- Using dilution formula we can determine the final molarity.
M1V1 = M2V2
M2 = M1V1 ÷ V2
= (12.0 M × 0.035 L) ÷ 1.2 L
= 0.35 M
Thus, the final concentration of the solution is 0.35 M
Answer:
27.9 g
Explanation:
CsF + XeF₆ → CsXeF₇
First we <u>convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles</u>, using its<em> molar mass</em>:
- Molar mass of CsXeF₇ = 397.193 g/mol
- 73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇
As <em>1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇</em>, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.
Now we <u>convert 0.184 moles of CsF to moles</u>, using the <em>molar mass of CsF</em>:
- Molar mass of CsF = 151.9 g/mol
- 0.184 mol * 151.9 g/mol = 27.9 g
