Answer:
Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
N₂ + O₂ ⟶ 2NO
N≡N + O=O ⟶ 2O-N=O
Bonds: 2N≡N 1O=O 2N-O + 2N=O
D/kJ·mol⁻¹: 941 495 201 607
(a)
The resistance of the rod is given by:
(1)
where
is the material resistivity
L = 1.20 m is the length of the rod
A is the cross-sectional area
The radius of the rod is half the diameter: , so the cross-sectional area is
The resistance at 20°C can be found by using Ohm's law. In fact, we know:
- The voltage at this temperature is V = 15.0 V
- The current at this temperature is I = 18.6 A
So, the resistance is
And now we can re-arrange the eq.(1) to solve for the resistivity:
(b)
First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is
I = 17.5 A
So the resistance is
The equation that gives the change in resistance as a function of the temperature is
where
is the resistance at the new temperature (92.0°C)
is the resistance at the original temperature (20.0°C)
is the temperature coefficient of resistivity
Solving the formula for , we find
An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.
Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.
This causes them to be soft and highly reactive because:
Examples of alkali electrons include:
In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.
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