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

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b