Chemistry, 07.07.2021 14:00, smithad1115

7. (a) Using two examples, explain how covalent compounds are formed. [4 marks] (b) State three physical properties of covalent compounds. [3 marks]
(c) Describe an experiment how you can carry out an experiment to investigate one of the properties mentioned above. [3 marks]

8. (a) Using two examples, explain how covalent compounds are formed. [14 marks]
(b) State three physical properties of covalent compounds.[3 marks]
Describe an experiment how you can carry out an experiment to investigate one of the properties mentioned above. [3 marks]

7. (a) Using two examples, explain how covalent compounds are formed. [4 marks]

(b) State three p

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answered: gabby0148

Answer; a) Covalent compound are formed when atoms of different molecules share electrons in order to be stable either by losing or gaining electrons

eg.. HCL hydrogen chloride , a chlorine atom has 7 electrons on its outer shell and hydrogen has 1 so when this two pair they become covalently bonded. chlorine shares one electron with the hydrogen atom ,thus to say both now have a stable arrangement of electron in their outer shell.

methane CH4 the carbon atom has 4 electrons and hydrogen has 1 so carbon would need 4 more and hydrogen would need 1 more so carbon pairs with 4 hydrogen atoms making it 8 and hydrogen 2 so they now have a stable arrangement ,thus the carbon atom shares with 4 hydrogen atoms.

b) some physical properties are

i)they have low melting and boiling points

ii)their insolubility in water

iii) they are non conductors of electricity

c) an experiment using their melting and boiling point

EXPERIMENT; lets use the covalent molecule carbon monoxide CO

when you apply heat and it tend to melt at a -199 degree celsius then it is a covalent molecule

and when it starts to boil at -191 degree celsuis then its sure a carbon monoxide covalent molecule

explanation; the reason for its low melting point and boiling point is because the attraction between the molecule is very low so it doesn't need much energy to break up the lattice between them.

HOPE THIS HELP if you need more explanation feel free to comment in the comment section

answered: Guest

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.