In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
Answer:
Group 7A
Explanation:
The group 7A elements consists of the most reactive non-metals on the periodic table.
This group is known as the group of halogens. They consist of element fluorine, chlorine, bromine, iodine and astatine.
- The elements in this group have the highest electronegativity values.
- They have 7 valence electrons and requires just one electron to complete their octets.
- This way, they are highly reactive in their search for that single electron.
Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:
As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used = 
so 1.54 moles of sodium azide will give = 
mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume = 
Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.
Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)
