Answer:
D no. is the answer of your question
An iron atom has a charge of zero. It's charge increases to 2+. Due to the increase in oxidation state, the atom is said to be oxidised. Oxidation occurs when an atom loses electrons.
Hence, the answer is (a).
Answer:
No component is perfect. All have tolerances that can vary. If you construct a simple circuit where a 10 volt power supply feeds a 10 ohm resistor, you would expect to measure a current of one ampere. BUT - the wiring has some resistance too. This adds perhaps 0.1 ohms to the circuit. The resistor has a +-5% tolerance. If it is 5% high, it may measure 10.5 ohms. That's a total circuit resistance of 10.6 ohms. The power supply may have a tolerance of +-1%. Suppose it's 1% low. That's an output of 9.9 volts in real life. So you have 9.9 volts dropped across 10.6 ohms. you will measure closer to 0.934 amps instead of 1.000 amps. To make matters worse, most electronic components have a temperature coefficient, that is, their values change with different temperatures. You may get a completely different reading tomorrow if the temperature is different! Finally, with current measurements in particular, you are inserting the ammeter in series with the circuit under test. Ammeters have some inherent resistance too, so by putting the ammeter in the circuit, you are changing the very current you are trying to measure (a little)! Oh yeah, the ammeter has a tolerance too. Its reading may be off a little even if everything else is perfect. Sometimes you have to wonder how we get a decent reading at all. Fortunately the errors are usually fairly small, and not all tolerances are off in the same direction or off the maximum amount. They tend to cancel each other out somewhat. BUT - in rare circumstances everything CAN happen like I said, and the error can be huge.
Explanation:
Change in Gibb's free energy of system (ΔG) = ΔH - TΔS.........(Eq. 1)
Now, if magnitude of ΔG <0, then reaction is spontaneous.
if magnitude of ΔG > 0, then reaction is non-spontaneous.
At equilibrium, ΔG = 0
When at boiling point, liquid state is in equilibrium with vapour state. Hence, it present case ΔG = 0
∴ Eq 1 becomes, ΔH = TΔS
here, ΔH = 58.2 kj/mol (Given),
∴ At T = 83.4 oC = 356.4 K, ΔS = 0.1633 kj/mol.K
