Answer:
1.32*10^23 molecules
Explanation:
sucrose formula: C12H22O11
molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol
75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11
0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)
The answer to this question is False
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:
a)
b)
Explanation:
a) The reaction:
The free-energy expression:
![E=E_{red}-E_{ox]](https://tex.z-dn.net/?f=E%3DE_%7Bred%7D-E_%7Box%5D)
The element wich is reduced is the Fe and the one that oxidates is the Mg:
The electrons transfered (n) in this reaction are 2, so:
b) If you have values of enthalpy and enthropy you can calculate the free-energy by:
with T in Kelvin
