Chemistry, 07.02.2021 01:50 Lilbre6999

Consider the following intermediate chemical equations. 2 equations. First; upper C upper H subscript 4 (g) plus 2 upper O subscript 2 (g) right arrow upper C upper O subscript 2 (g) plus 2 upper H subscript 2 upper O (g). Second: 2 upper H subscript 2 upper O (g) right arrow 2 upper H subscript 2 upper O (l).

Which overall chemical equation is obtained by combining these intermediate equations?
Upper C upper H subscript 4 plus 2 upper O subscript 2 (g) right arrow upper C upper O subscript 2 (g) plus 2 upper H subscript 2 upper O (l).
Upper C upper H subscript 4 plus 2 upper O subscript 2 (g) right arrow upper C upper O subscript 2 (g) plus 2 upper H subscript 2 upper O (g).
Upper C upper H subscript 4 plus 2 upper O subscript 2 (g) right arrow upper C upper O subscript 2 (g) plus 4 upper H subscript 2 upper O (g) plus 2 upper H subscript 2 upper O (l).
Upper C upper H subscript 4 plus 2 upper O subscript 2 (g) right arrow upper C upper O subscript 2 (g) plus 6 upper H subscript 2 upper O (g).

Answers: 1

Show answers

Answers

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

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.