Answer: none of the above
Explanation:
It should be 2,1,1,2 to give a balanced chemical reaction
<u>We are given:</u>
M1 = 3 Molar V1 = 80 mL
M2 = x Molar V2 = 100 mL
<u>Finding the molarity:</u>
We know that:
M₁V₁ = M₂V₂
where V can be in any units
(3)(80) = (x)(100)
x = 240/100 [dividing both sides by 100]
x = 2.4 Molar
Answer:
![\delta H_{rxn} = -66.0 \ kJ/mole](https://tex.z-dn.net/?f=%5Cdelta%20H_%7Brxn%7D%20%3D%20-66.0%20%20%5C%20kJ%2Fmole)
Explanation:
Given that:
![3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)](https://tex.z-dn.net/?f=3FeO_3_%7B%28s%29%7D%2BCO_%7B%28g%29%7D%20%5Cto%202Fe_3O_4_%7B%28s%29%7D%20%2BCO_%7B2%28g%29%7D%20%5C%20%20%5C%20%5Cdelta%20H%20%3D%20-47.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%281%29%20%20%5C%5C%20%5C%5C%20%5C%5C%20Fe_2O_3_%7B%28s%29%7D%20%2B3CO_%7B%28g%29%7D%20%5Cto%202FE_%7B%28s%29%7D%20%2B%203CO_%7B2%28g%29%7D%20%20%5C%20%5C%20%5Cdelta%20H%20%3D%20-25.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%282%29%20%20%5C%5C%20%5C%5C%20%5C%5C%20Fe_3O_4_%7B%28s%29%7D%20%2B%20CO_%7B%28g%29%7D%20%5Cto%203FeO_%7B%28s%29%7D%20%2B%20CO_%7B2%28g%29%7D%20%5C%20%5Cdelta%20H%20%3D%2019.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%283%29)
From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:
![3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)](https://tex.z-dn.net/?f=3FeO_%7B%28s%29%7D%20%2B%20CO_%7B2%28g%29%7D%20%20%20%20%5Cto%20%20%20%20Fe_3O_4_%7B%28s%29%7D%20%2B%20CO_%7B%28g%29%7D%20%20%20%5C%20%5Cdelta%20H%20%3D%20-19.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%284%29)
Multiplying (2) with equation (4) ; we have:
![6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)](https://tex.z-dn.net/?f=6FeO_%7B%28s%29%7D%20%2B%202CO_%7B2%28g%29%7D%20%20%20%20%5Cto%20%20%20%202Fe_3O_4_%7B%28s%29%7D%20%2B%202CO_%7B%28g%29%7D%20%20%20%5C%20%5Cdelta%20H%20%3D%20-38.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%285%29)
From equation (1) ; multiplying (-1) with equation (1); we have:
![2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)](https://tex.z-dn.net/?f=2Fe_3O_4_%7B%28s%29%7D%20%2BCO_%7B2%28g%29%7D%20%5Cto%20%20%20%20%203FeO_3_%7B%28s%29%7D%2BCO_%7B%28g%29%7D%20%20%20%5C%20%20%5C%20%5Cdelta%20H%20%3D%2047.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%286%29)
From equation (2); multiplying (3) with equation (2); we have:
![3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)](https://tex.z-dn.net/?f=3%20Fe_2O_3_%7B%28s%29%7D%20%2B9CO_%7B%28g%29%7D%20%5Cto%206FE_%7B%28s%29%7D%20%2B%209CO_%7B2%28g%29%7D%20%20%5C%20%5C%20%5Cdelta%20H%20%3D%20-75.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%287%29)
Now; Adding up equation (5), (6) & (7) ; we get:
![6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)](https://tex.z-dn.net/?f=6FeO_%7B%28s%29%7D%20%2B%202CO_%7B2%28g%29%7D%20%20%20%20%5Cto%20%20%20%202Fe_3O_4_%7B%28s%29%7D%20%2B%202CO_%7B%28g%29%7D%20%20%20%5C%20%5Cdelta%20H%20%3D%20-38.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%285%29)
![2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)](https://tex.z-dn.net/?f=2Fe_3O_4_%7B%28s%29%7D%20%2BCO_%7B2%28g%29%7D%20%5Cto%20%20%20%20%203FeO_3_%7B%28s%29%7D%2BCO_%7B%28g%29%7D%20%20%20%5C%20%20%5C%20%5Cdelta%20H%20%3D%2047.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%286%29)
![3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)](https://tex.z-dn.net/?f=3%20Fe_2O_3_%7B%28s%29%7D%20%2B9CO_%7B%28g%29%7D%20%5Cto%206FE_%7B%28s%29%7D%20%2B%209CO_%7B2%28g%29%7D%20%20%5C%20%5C%20%5Cdelta%20H%20%3D%20-75.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%287%29)
<u> </u>
![FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole](https://tex.z-dn.net/?f=FeO%20%20%5C%20%5C%20%5C%20%2B%20%20%5C%20%5C%20%5C%20CO%20%20%20%5C%20%5C%20%20%5Cto%20%20%20%5C%20%5C%20%5C%20%5C%20Fe_%7B%28s%29%7D%20%2B%20%5C%20%5C%20CO_%7B2%28g%29%7D%20%5C%20%5C%20%5C%20%20%5Cdelta%20H%20%3D%20-%2066.0%20%5C%20kJ%2Fmole)
<u> </u>
<u />
(According to Hess Law)
![\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole](https://tex.z-dn.net/?f=%5Cdelta%20H_%7Brxn%7D%20%3D%20%28-38.0%20%2B%20%2047.0%20%2B%20%28-75.0%29%29%20%5C%20kJ%2Fmole)
![\delta H_{rxn} = -66.0 \ kJ/mole](https://tex.z-dn.net/?f=%5Cdelta%20H_%7Brxn%7D%20%3D%20-66.0%20%20%5C%20kJ%2Fmole)
The answer is A: Between 50 and 5,000 amino acids
Answer:
C.
Explanation:
It’s particle boiling point because the atoms are moving fast around the particle as possible, so there for its C..