The best name for the ionic bond that forms between them is Beryllium Bromide.
We have been provided with data,
Beryllium charge, q = 2
Bromine charge, q = -1
As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.
Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.
So, the best name for the ionic bond that forms between them is Beryllium Bromide.
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Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
Ya know what the day you are talking to you about it and then send me the back link you got it and I don’t want
Answer:
The first graph is showing the constant acceleration (1 m/s)
Explanation:
The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4
The last graph is showing constant velocity therefore there is no acceleration (a = 0)
