Answer:
14.32g
Explanation:
Initial temperature = 83.8°C
Final temperature = 77.1°C
Temperature change, ΔT = 83.8°C - 77.1°C = 6.7
Heat, H = 167J
Specific heat, c = 1.740J/g °C
m = ?
All these parameters are related with the equation below;
H = mcΔT
m = H / cΔT
m = 167 / (1.740 * 6.7)
m = 167 / 11.658 = 14.32g
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
100. Centi means a hundred.
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Answer : The new pressure if the sample placed in a 2.00-L container is, 0.494 atm.
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

or,

where,
= initial pressure = 0.988 atm
= final pressure = ?
= initial volume = 1.00 L
= final volume = 2.00 L
Now put all the given values in the above equation, we get:


Thus, the new pressure if the sample placed in a 2.00-L container is, 0.494 atm.