How do the number of electrons in the second energy shell of an atom change going across period 2 in the periodic table? a. the number of electrons increases by 1, from left to right. b. atoms in period 2 do not have a second energy shell, instead the first energy shell increases in electrons from left to right. c. the number of electrons in the second energy shell is full, the number of electrons in the third energy shell increase by 1. d. the number of electrons increases by 1, from right to left.

The answer should be A, the number of electrons increases by 1, from left to right.

Across the periods (horizontal direction) in periodic table, each element has one more electron then the one before it. The number of period determines the number of occupied electron shells of the elements in that period. Which means, period 1 elements have 1 occupied electron shell, period 2 has 2, period 3 has 3 etc.

Therefore, the answer is A, we always go from left to right, and plus since period 2 elements only have 2 occupied electrons so their 2nd electron shell electrons must increase throughout the period.

Answer : option a) 84explanation :

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.

The reaction 2kalsi3o8 + 2(h+ + hco3-) + h2o  →  al2si2o5(oh)4 + 2k+ + 2hco3- + 4sio2 shows that feldspar kalsi3o8 is chemically altered by hydrolysis to form chalky-looking clay mineral kaolinite al2si2o5(oh)4. the chemical breakdown of potassium feldspar due to exposure to acidic water results to the mineral binding with the water molecule to become kaolinite with silica and potassium and bicarbonate ions in solution.