Answer: Do you take Ms. Reyes
Explanation: The answer is Outside air, Nose, Lungs, Bloodstream, Cell
Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%5BOH%5E-%5D%7D)
The concentration of OH⁻ is:
pOH = 14 - pH = <em>3</em>
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%5B1x10%5E%7B-3%7D%5D%7D)
<em>0,04M = ![\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAl%28OH%29_%7B2%7D%20%5E-%5D%7D%7B%5BAl%28OH%29%5D%7D)
</em>
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
Answer:
43.2 moles of carbon dioxide are required and 421g of glucose could be produced
Explanation:
Based on the reaction:
6CO2 + 6H2O → C6H12O6 + 6O2
1 mole of glucose, C6H12O6, requires 6 moles of carbon dioxide. 7.2moles of glucose requires:
7.2mol C6H12O6 * (6mol CO2 / 1mol C6H12O6) =
<h3>43.2 moles of carbon dioxide are required</h3><h3 />
618g of CO2 -Molar mass: 44.01g/mol- are:
618g * (1mol / 44.01g) = 14.04moles CO2
Moles C6H12O6:
14.04moles CO2 * (1mol C6H12O6 / 6mol CO2) = 2.34moles C6H12O6
Mass glucose -Molar mass: 180.156g/mol-
2.34moles C6H12O6 * (180.156g / mol) =
<h3>421g of glucose could be produced</h3>
Liquid is water or drink and solid is hard
Answer: 2 atoms
Explanation: A molecular formula of the compound carbon dioxide is CO 2 . One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen.
