A neutral atom has a configuration of 1s22s22p5. Which group on the periodic table contains this atom?
A. 3
B. 5
C. 17
D. 2

C

Explanation:

The atom has an electron configuration of 1s²2s²2p⁶3s²3p⁵.

The total numbe of electrons present in the shell is 17 electrons. To determine the group of the atom, we have to check what orbital did the last electron stop.

If it's in S-orbital, then we automatically knows it's an s-block element which could either be group 1 or 2 and can still be furthered down to know which group it's in.

But in this case, the last electron is in s-orbital. To determine the group number of p-orbitals, we'll have to consider the preceding s-orbital and add it to it, then count it in 10s.

For example, if the last two orbital is 3s² 3p¹, 2 + 1 = 3, add ten (10) to it = 13. Hence the element is in group 13.

In this case, we have 3s² 3p⁵ = 2 + 5 = 7 + 10 = 17.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :