Answer:
136 g Al₂O₃
Explanation:
Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).
Molar Mass (O₂): 32 g/mol
Molar Mass (Al₂O₃): 102 g/mol
4 Al + 3 O₂ -----> 2 Al₂O₃
64.0 g O₂ 1 mole 2 moles Al₂O₃ 102 g
----------------- x -------------- x ------------------------ x ------------- = 136 g Al₂O₃
32 g 3 moles O₂ 1 mole
Answer:
K=CHANGE IN CONCENTRATION/TIME TAKEN
Explanation:
0.2 g/mL
We can use the density formula <em>D = M/V</em>.
<em>D</em> = 6 g/30.0 mL = 0.2 g/mL
Answer:
- Mass of NaH₂PO₄·H₂O = 8.542 g
- Mass of Na₂HPO₄ = 5.410 g
Explanation:
Keeping in mind the equilibrium:
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺
We use the Henderson-Hasselbalch equation (H-H):
pH = pka + ![log\frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]
From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that
[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M
From this equation we can <u>express [H₂PO₄⁻] in terms of [HPO₄⁻²]</u>:
[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]
And then replace [H₂PO₄⁻] in the H-H equation, <u>in order to calculate [HPO₄⁻²]</u>:
![7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]](https://tex.z-dn.net/?f=7.00%3D7.21%2Blog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%5D%20%7D%7B0.100%20M-%5BHPO4%5E%7B-2%7D%5D%7D%20%5C%5C-0.21%3Dlog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%5D%20%7D%7B0.100%20M-%5BHPO4%5E%7B-2%7D%5D%7D%5C%5C10%5E%7B-0.21%7D%20%3D%5Cfrac%7B%5BHPO4%5E%7B-2%7D%5D%20%7D%7B0.100%20M-%5BHPO4%5E%7B-2%7D%5D%7D%5C%5C0.616%2A%280.100M-%5BHPO4%5E%7B-2%7D%5D%29%3D%5BHPO4%5E%7B-2%7D%5D%5C%5C0.0616%20M%20%3D%201.616%2A%5BHPO4%5E%7B-2%7D%5D%5C%5C0.03812%20M%20%3D%5BHPO4%5E%7B-2%7D%5D)
With the value of [H₂PO₄⁻],<u> we calculate [HPO₄⁻²]</u>:
[HPO₄⁻²] + 0.0381 M = 0.100 M
[HPO₄⁻²] = 0.0619 M
Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:
- Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g
- Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g
Answer:
The mitochondria change energy in organic compounds into a
form the cell can use.
The chloroplasts take in energy from sunlight and change it into
organic matter.
Explanation: