Most likely the equation yields a new compound XY. This is a synthesis reaction. <span>Synthesis
reaction is a reaction where two or more substances combine to form a new
compound. It is a reaction which releases energy in the form of heat and
light. Therefore, it is an exothermic reaction.</span>
Answer: mass : kg, length: meter, time: second, temperature: Kelvin
Explanation:
Mass is defined as the amount of matter contained in the body.
Its units are kg, gram, milligram which are inter convertible.
S.I or M.K.S system has seven fundamental units which are used to find derived units
1) Mass - Kilogram
2) Length - meter
3) Time - Seconds
4) Electric Current - Ampere
5) Amount of substance - Moles
6) Intensity of light - Candela
7) Temperature - Kelvin
Thus SI base unit of each of these quantities are kg, meter, second, and Kelvin
41. Mercury,Venus, Earth,mars,Jupiter, Saturn,Uranus and Neptune
42. Nebula - Their birth places are huge, cold clouds of gas and dust
As given that some volume of water has been dispensed say "x mL"
The initial weight of bottle =8.4376 g
The final weight of bottle + water =28.5845 g
So weight of water transferred = 28.5845 g - 8.4376 g = 20.1469 g
Now there is a relation between density, mass and volume
density = mass / volume
Therefore
Volume = mass / density
So volume of water dispensed = mass dispensed / density =20.1469 g / 0.9967867 g/ml.
Volume of water = 20.2118467 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