Actually Rb or Rubidium in zero state has the following
electron configuration:
<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>
However we can see that the ion has a 1 positive charge,
which means that it lacks 1 electron, therefore the answer from the choices is:
<span>d.
rb+: 1s22s22p63s23p64s23d104p6</span>
It’s c had this problem last week
<span>1. </span>To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 x V1 / P2
V2 = 203 x 40.0 / 35.0
V2 =232 L
