The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Because the proton has a mass of 1 and a neutron has a mass of 1, we know that there is exactly one proton in the nucleus (because of the atomic number) which therefore tells us there are no neutrons as adding one would make the mass more than one.
Answer
A *single covalent bond* is formed by sharing one pair of valence electrons.They are less reactive comparatviely and have a high bond length
A *double covalent bond* is formed by sharing two pairs of valence electrons.They are moderately reactive and have moderate bond length.
A *triple covalent bond* is formed by sharing three pairs of valence electrons.They are highly reactive and have a low bond length.
Answer:
The sharing of electrons between a water molecule that forms four hydrogen bonds with the other four water molecules:
Explanation:
The hydrogen bond is a weak electrostatic force of attraction that exists between a covalently bonded H-atom and a highly electronegative atom like N,O or F.
In the case of the water molecule,
the highly electronegative atom is Oxygen and the intermolecular hydrogen bond in water is as shown below:
Thus H-bond is a weak electrostatic attraction formed between H-atom and O-atom in water.
Answer:
7.5 moles of CaBr2 are produced
Explanation:
Based on the equation:
2AlBr3 + 3CaO → Al2O3 + 3CaBr2
<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>
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Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:
5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =
<h3>7.5 moles of CaBr2 are produced</h3>
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