If a pork roast must absorb 1700 kj to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of co2 is emitted into the atmosphere during the grilling of the pork roast?

The amount of heat needed to cook the pork roast is 1700 kJ.

The amount of heat absorbed by the roast is 14% of the heat produced  = 14% X ?H = 310.38 kJ/mol

One mole of propane burns to produce 3 moles of CO2 and -2217 kJ of heat.

1700kJ/ (14% X ?H) = mol of C3H8 burned

1700kJ / (310.38 kJ/mol) = 5.48 mol C3H8

Since 3 moles of CO2 are produced:

3 X 5.48 mol= 16.43 mols of CO2

convert moles to grams:

16.43 mol CO2 X 44.0g/mol ~ 722.9 g CO2 produced

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12