Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)
Answer:
Iodine, Silicon, Copper.
Explanation:
Iodine is an insulator, which doesn't conduct electricity. Silicon is a semiconductor. Copper is a conductor.
Here’s what I found:
It takes very little energy to remove that outermost electron from an alkali metal. Thus, alkali metals easily lose their outermost electron to become a +1 ion. ... In fact, as you go down the 1A column, the first ionization energies get lower and lower, making cesium the most easily ionized element on the periodic table.
So basically it’s because part of what makes alkali metals so reactive is that they have one electron in their outermost electron layer.
PV=nRT
PV= (m/M)RT
PM=(m/v)RT
PM =dRT
d= (PM) ÷(RT)
substitute the the given data to the above equation. d-density, p-pressure, M- molar mass of argon gas, T- temperature. and R is a constant. make sure to substitute its value too.. :)
They are more likely to be covalent as covalent compounds usually tend to be radical on the color spectrum with deep purples and blacks and even earwax colored.
