This statement is false. C6H12O6 is not a diatomic molecule. A diatomic molecule has only two atoms. Diatomic molecules are molecule that are composed of two same atoms which are bonded covalently. An example are oxygen, it exist as O2 and chlorine, it exist as Cl2.
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!
Answer:
The percent composition is 21% N, 6% H, 24% S and 49% O.
Explanation:
1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.
In 1 mole of (NH4)2SO4, there are:
- 2 moles of N.
- 8 moles of H.
- 1 mole of S.
- 4 moles of O.
2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:
- 2 moles of N (14g/mol) = 28g
- 8 moles of H (1g/mol) = 8g
- 1 mole of S (32g/mol) = 32g
- 4 moles of O (16g/mol) = 64g
3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:
In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.
So, the percent composition is 21% N, 6% H, 24% S and 49% O.
Answer:
Explanation:
a physical change is the change of the appearance, and a chemical change is the change of the thing.
D . on orbits or energy levels around the nucleus