Chemistry, 30.06.2019 14:30 kyriebarnes143

What types of atomic orbitals are in the third principal energy level? p and d only s, p, d, and f s, p, and d only

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Answer from: emmdolzx

- s, p and d only.

Explanations:- Principal energy level or quantum number is represented by n which is an integer number. n has values like 1,2,3 and so on.

For first shell, the value of n is 1, for second shell the value of n is 2, the value of n for third shell is 3 and so on.

Principal energy levels are divided into sub energy levels. First principal energy level or shell has s sub energy level or sub shell, second shell has s and p sub shells, third shell has s, p and d sub shells and similarly the fourth shell has s, p, d and f sub shells.

Azimuthal quantum number tell us about the sub shells. Azimuthal quantum number is represented by l and it's values depends on the values of n.

values of l are 0 to n-1. For example, for the first shell, n = 1 then l = 0.

for second shell, n = 2 then l = 0 and 1.

for third shell, n = 3 and l = 0, 1 and 2.

l = 0 means s sub shell, l = 1 means p sub shell and l = 3 means d sub shells and so on.

Since, the value of n for third principle energy level(shell) is 3 and the values of l for this value of n are 0, 1 and 2. The sub shells are s, p and d and same are the alphabets used for the orbitals.

So, the right choice is s, p and d only.

Answer from: Quest

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.

Answer from: Quest

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

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

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.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

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

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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What types of atomic orbitals are in the third principal energy level? p and d only s, p, d, and f...

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