Chemistry, 22.12.2019 19:31 picklespicklesdicc

Which of the following is the trend down a period?
a. the number of complete electron shells increases by one.
b. the number of complete electron shells stays the same.
c. the number of outer shell electrons decreases by one.
d. none of these.

Answers: 1

Show answers

Answers

Answer from: marabic22

B. The number of complete electron shells stays the same.

Explanation:

The periods on a periodic table consists of those elements that are arranged in the horizontal row in the periodic table.

Elements in the same period have the same number of electron shells and this corresponds to the period. This implies that elements of period 2 have two electron shells, those of period 3 have three shells and so on.

The number of Valence electrons of the elements on the same period increases progressively by one across the period from left to right.

A group is the vertical column to which elements are arranged on the periodic table. Elements in the same group have the same number of Valence electrons and the number of shell increases down a group.

When we go down a period, the number of shell stays the same.

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 :

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

Answer from: Quest

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.