Consider the reaction ca(oh)2(s)→cao(s)+h2o(l) with enthalpy of reaction δhrxn∘=65.2kj/mol what is the enthalpy of formation of cao(s)? express your answer in kilojoules per mole to one deci

ΔH °fCaO= -655.09 KJ/mol =  ΔH °fCaO

Explanation:

ΔH °(reaction)a=ΔH °f(product)-ΔH °f(reactant)

where,

ΔH °f(product)=is the standard heat of formation of products

-ΔH °f(reactant)=is the standard heat of formation of reactants

The standard heat of formation of  H 2 O ( l )  is  − 285.8 k J / m o l .

The standard heat of formation of  Ca(OH )2  is  − 986.09 k J / m o l .

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

ΔH °(reaction)a=ΔH °f(CaO+H2O)-ΔH °f(Ca(OH)₂

65.2                =  ΔH °fCaO+( -265.8) -(-986.09)

                       = ΔH °fCaO-265.8+986.09

65.2                        =  ΔH °fCaO+ 720.29

65.2-720.29      =  ΔH °fCaO

-655.09 KJ/mol ==  ΔH °fCaO

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.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3