Answer:
its the lower brainstem
Explanation:
The lowest part of the brainstem, the medulla is the most vital part of the entire brain and contains important control centers for the heart and lungs.
Answer:
K = 2.96x10⁻¹⁰
Explanation:
Based on the initial reaction:
N2O4 ⇄ 2NO2; K = 1.5x10³
Using Hess's law, we can multiply this reaction changing K:
3 times this reaction:
3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹
The inverse reaction has a K of:
6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;
<h3>K = 2.96x10⁻¹⁰</h3>
Answer:
carbon dioxide
Explanation:
Carbon burns in oxygen to form carbon dioxide. Since hydrocarbon fuels only contain two elements, we always obtain the same two products when they burn. In the equation below methane (CH 4) is being burned. The oxygen will combine with the carbon and the hydrogen in the methane molecule to produce carbon dioxide (CO 2) and water (H 2O).
Carbon, as graphite, burns to form gaseous carbon (IV) oxide (carbon dioxide), CO2. ... When the air or oxygen supply is restricted, incomplete combustion to carbon monoxide, CO, occurs. 2C(s) + O2(g) → 2CO(g) This reaction is important. When one mole of carbon is exposed to some energy in the presence of one mole of oxygen gas, one mole of carbon dioxide gas is produced. This reaction is a combustion reaction.
Answer:
s = 4.41 g/L.
Explanation:
¡Hola!
En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

Lo cual hace que la expresión de equilibrio se calcule como:
![Ksp=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
Y que en términos de la solubilidad molar, s, se resuelve como:
![1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L](https://tex.z-dn.net/?f=1.6x10%5E%7B-5%7D%3Ds%282s%29%5E2%5C%5C%5C%5C1.6x10%5E%7B-5%7D%3D4s%5E3%5C%5C%5C%5Cs%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1.6x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cs%3D0.0159molPbCl_2%2FL)
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

¡Saludos!