Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer:
The enthalpy of the reaction is 64.9 kJ/mol.
Explanation:
..[1]
..[2]
..[3]
..[4]
2 × [2] - [1]- [3] = [4] (Using Hess's law)
The enthalpy of the reaction is 64.9 kJ/mol.
Answer:
The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode
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