Answer:
Rate of appearance of Bromine is 2.04 X 10-4 M·s−1
Explanation:
According to the balanced stoichiometric equation, for every 5 moles of Br- that react, 3 moles of Br2 is produced so they have the stoichiometric ratio of 5:3.
We can use this to calculate the rate of appearance of bromine since this is a direct relationship.
Br- : Br2
5 : 3
3.4 ✕ 10−4 M·s−1 : x
5x = 1.02 X 10-3 M·s−1
x = 2.04 X 10-4 M·s−1
Answer:
![9.09*10^{21} molecules O_2](https://tex.z-dn.net/?f=9.09%2A10%5E%7B21%7D%20molecules%20O_2)
Explanation:
The formula for Mercury(II) Oxide is ![HgO](https://tex.z-dn.net/?f=HgO)
Balanced Equation: ![2HgO = 2Hg + O_2](https://tex.z-dn.net/?f=2HgO%20%3D%202Hg%20%2B%20O_2)
Stoichiometry: ![(6.54g HgO)*(\frac{1molHgO}{216.59gHgO})*(\frac{1molO_2}{2molHgO})*(\frac{6.02*10^{23}molecules O_2}{1molO_2})= 9.09*10^{21}moleculesO_2](https://tex.z-dn.net/?f=%286.54g%20HgO%29%2A%28%5Cfrac%7B1molHgO%7D%7B216.59gHgO%7D%29%2A%28%5Cfrac%7B1molO_2%7D%7B2molHgO%7D%29%2A%28%5Cfrac%7B6.02%2A10%5E%7B23%7Dmolecules%20O_2%7D%7B1molO_2%7D%29%3D%209.09%2A10%5E%7B21%7DmoleculesO_2)
I believe the first one is chemical, second one is mechanical, third one is mechanical and fourth one is chemical.
Answer:
The larger piston
Explanation:
(i just took the same test)
2NH3 + H2O2 → N2H4 + 2H2O