4. The equation for the decomposition of hydrogen peroxide is shown. 2H₂O₂(aq) → 2H₂O(l) + O₂(g) 48.0 25.0 cm³ of aqueous hydrog
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The empirical and molecular formulas will be and
respectively.
The compound contains C, H, and O.
C = 61.15/12 = 5.0958
H = 5.3/1 = 5.3
O = 31.55/16 = 1.9719
Divide by the smallest
C = 2.6
H = 2.7
O = 1
Thus, the empirical formula is
Empirical formula mass = (12x5) + (1x5) + 16x2 = 97
n = 152.15/97 = 2
The molecular formula is
More on molecular and empirical formulas can be found here: brainly.com/question/14425592
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Answer:
1. 0.97 V
2.
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!