There are two sides on a relay, the coil side for activating the relay, and the contact side, for switching.
Your relay is a 5 V relay, that's the activating side. If you look at the (messy) datasheet you'll see that it has a resistance of 55 Ω, for a 450 mW power. Some of that power is used to create the magnetic field needed to close the contact, but most is dissipated in the coil's resistance as heat. Power can be calculated as
\$ P = \dfrac{V^2}{R} \$
and at 5 V that's indeed 450 mW. If you would power it with 120 V then
\$ P = \dfrac{(120 V)^2}{55 \Omega} = 260 W\$
will most likely will make the relay explode. The datasheet says the maximum voltage is 110 % of the rated voltage, so that's 5.5 V maximum.
That activation power will switch any voltage and current as long as they're below the rated values. You don't need a higher activation voltage to switch 10 A than for 1 A, or for 120 V instead of 12 V.
The contacts are rated at 10 A, for DC a maximum voltage of 30 V is given for the form A contact (SPST). This is a form C contact (SPDT) however, and this doesn't specify a DC rating for a resistive load. For AC it's 240 V. (JeeShen's answer is not correct here.) But power is maximum 240 W, so at 120 V AC you can only switch a resistive load of 2 A.
Conservative engineering practices, and my own "school of hard knocks" experience say that you have to size the relay for the inrush current, because even a brief surge can weld the contacts, in which case the relay is stuck on and could lead to a bad safety problem. But don't decide by looking at the brief ratings in a catalog. Look at the data sheet and see if there is an inrush rating that is higher than the continuous duty rating. Motor starting contactors are an extreme example of this. They have contact materials that resist welding, and their ratings reflect this.
Another approach that you see in large power supplies is a separate set of contacts, in series with a power resistor, that are closed for a second or so before the main relay closes, to more gradually charge the filter capacitors. Sometimes referred to as a "soft start" circuit.
Best Answer
The most common cause of relay chatter is the supply to the relay coil falling below coil hold on point...
this relaxes the load on the supply....
The supply then rises again above the coil pull-in point...
The supply sags again and the cycle repeats.
The solution in most cases is a higher current supply.
EDIT...... Sometimes it is not the supply current that is the problem but a high resistance between the coil and the supply. This could be because the driving transistor or FET not being driven hard enough or a bad connection.
Examples are dry solder joints or loose screw terminals
EDIT 2...
Also this rapid switching can introduce noise in to the supply that can affect sensor circuits. The supply rail going up and down will affect circuits that assume a stable supply rail. This could cause the wrong action to be taken by the micro (corrupt inputs) and in bad cases the micro to reset depending on the circuit.
The contacts opening and closed rapidly increases the production of EMI as well, each arc of the contacts sending pulses in to the environment and the rest of the circuit ...