Answer:
The empirical formula for C12 H24 O6 is C2 H4 O.
Answer:
CaS, CaBr₂, VBr₅, and V₂S₅.
Explanation:
- The ionic compound should be neutral; the overall charge of it is equal to zero.
- Binary ionic compound is composed of two different ions.
<u>Ca²⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- CaS can be formed via combining Ca²⁺ with S²⁻ to form the neutral binary ionic compound CaS.
- CaBr₂ can be formed via combining 1 mole of Ca²⁺ with 2 moles of Br⁻ to form the neutral binary ionic compound CaBr₂.
<u>V⁵⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- V₂S₅ can be formed via combining 2 moles of V⁵⁺ with 5 moles of S²⁻ to form the neutral binary ionic compound V₂S₅.
- VBr₅ can be formed via combining 1 mole of V⁵⁺ with 5 moles of Br⁻ to form the neutral binary ionic compound VBr₅.
<em>So, the empirical formula of four binary ionic compounds that could be formed is: CaS, CaBr₂, VBr₅, and V₂S₅.</em>
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Answer:
The final volume of the sample of gas 
= 0.000151 
Explanation:
Initial volume 
= 200 ml = 0.0002
Initial temperature 
= 296 K
Initial pressure 
= 101.3 K pa
Final temperature 
= 336 K
Final pressure 
= K pa
Relation between P , V & T is given by
Put all the values in the above equation we get
= 0.000151
This is the final volume of the sample of gas.
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
