For the answer to the question above, well presumably because the exact concentration of the composition KMnO4 solution doesn't matter. <span>If the concentration of the KMnO4 solution is important (usually in titrations etc.) then it is not allowed to use a wet bottle. The water in the bottle will dilute the KMnO4 solution and change the concentration of the said compound.</span>
Answer:
32.7 g of Zn
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Zn + 2HCl —> ZnCl₂ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂
Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.
Thus, 0.5 mole of Zn is required.
Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:
Mole of Zn = 0.5 mole
Molar mass of Zn = 65.4 g/mol
Mass of Zn =?
Mass = mole × molar mass
Mass of Zn = 0.5 × 65.4
Mass of Zn = 32.7 g
Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.
Answer:
zince chloride will be formed and hydrogen gas will be librated.
Explanation:
When dilute HCl is added to zinc pieces, a rection will takes place as follows :

It means that zinc chloride will form when zinc reacts with dilute HCL. Also hydrogen gas will produced.
As zinc is more reactive than hydrogen, it displaces hydrogen from its solution and forms zinc chloride. The form product is white in color and H₂ is an odorless gas.
Hence, zince chloride will be formed and hydrogen gas will be librated.
At least, that's what Bohr<span> decided, and that's why he proposed the </span>existence<span> of the</span>atomic<span> energy level. </span>According<span> to </span>Bohr<span>, the electrons in an </span>atom<span> were only allowed to </span>exist<span> at certain energy levels</span>
In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.
O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13
CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044
Answers: O2 = 0.13
CO2 = 0.044