Let us assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2
heat absorbed by pork = 0.11 x 2217
= 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
= 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
= 44 x 20.913
= 920.172g of CO2 emitted
Answer:
B I believe im sorry if it's not right
Answer:
from south to north thats what it look like
Explanation:
Rare earth elements are a series of chemical elements found in the earth's crust and are vital to many of the modern technologies in the world such as computers and networks, advanced transportation and consumer electronics. They help fuel economic growth, maintain high living standards and even save lives. Examples include:
Scandium. Used in television and fluorescent lamps.
Yttrium. Used in cancer treatment drugs, superconductors and camera lenses
Lanthanum. Used to make special optical glasses, telescope lenses and also in petroleum refining.
Neodymium. Used in making some of the strongest permanent magnets, found in most modern vehicles and aircraft.
Answer:
0.0119
Explanation:
There was a part missing. I think this is the whole question:
<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>
A (aq) ⇌ 2B (aq) + C(aq)
<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>
First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.
A (aq) ⇌ 2B (aq) + C (aq)
I 0.0510 0 0
C -x +2x +x
E 0.0510-x 2x x
Since the concentration at equilibrium of A is 0.0153 M, we get

We can use the value of x to calculate the concentrations at equilibrium.
![[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\](https://tex.z-dn.net/?f=%5BA%5De%20%3D%200.0153%20M%20%5C%5C%5BB%5De%20%3D%202x%20%3D%202%280.0357%29%20%3D%200.0714%20M%20%5C%5C%5BC%5De%20%3D%20x%20%3D%200.0357%20M%20%5C%5C)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
![Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BB%5D%5E%7B2%7D%20%20%5Ctimes%20%5BC%5D%7D%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B0.0714%5E%7B2%7D%20%20%5Ctimes%200.0357%7D%7B0.0153%7D%20%3D%200.0119)
The equilibrium constant for this reaction at equilibrium is 0.0119.
You can learn more about equilibrium here: brainly.com/question/4289021