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 least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation:

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!