To solve for the empirical formula, we write first all the data.
Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8 wt% O
Required: Empirical Formula of Compound 1
Solution:
Assume total mass of the compound is 100 g
Solving for Compound 1,
76 g Ru x <u>1 mol Ru </u> = 0.75195 mol Ru
101.07 g Ru
24 g O x <u>1 mol O </u> = 1.5 mol O
16 g O
Then, divide each mole with the smallest number of moles calculated
Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2
Therefore, the empirical formula for Compound 1 is RuO2.
<em>ANSWER: RuO2</em>
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Answer: 1000x
Explanation:
I hope this helped!
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- Zack Slocum
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1. 0+15+30+45+60=150/5=30 s
0+10+16+22+27=75/5=15 cm^3
then, 15/30= 0.5 cm^3/s(average rate of rxn)
2. 0+15+30+45+60+75+90+105+120=540/9=60 s
0+10+16+22+27+31.50+36+39.5+42=224/9=24.89cm^3
then, 24.89/60=0.414cm^3/s (avrg rate of rxn)
3. overall rxn add all the time divide by 12 and the volume add them too and divide by 12. after take average volume divide by the average time to get the average rate of overall rxn
4. 16/30= 0.533cm^3/s
5. 39.5/105= 0.37cm^3/s
Answer:
Aluminum sulfate with barium nitrate to produce burin and aluminum nitrate
