Answer:
8.33 hours
Explanation:
In order to solve this problem, we must apply Graham's law of diffusion in gases. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. For two gases we can write;
R1/R2=√d2/d1
Where;
R1= rate of diffusion of hydrogen
R2= rate diffusion of unknown gas
d1= vapour density of hydrogen
d2= vapour density of the unknown gas
Volume of hydrogen gas = 360cm^3
Time taken for hydrogen gas to diffuse= 1 hour =3600 secs
R1 = 360 cm^3/3600 secs = 0.1 cm^3 s-1
Vapour density of unknown gas = 25
Vapour density of hydrogen = 1
Substituting values,
0.1/R2 = √25/1
0.1/R2 = 5/1
5R2 = 0.1 × 1
R2 = 0.1/5
R2= 0.02 cm^3s-1
Volume of unknown gas = 600cm^3
Time taken for unknown gas to diffuse= volume of unknown gas/ rate of diffusion of unknown gas
Time taken for unknown gas to diffuse= 600/0.02
Time= 30,000 seconds or 8.33 hours
Answer:
52.8 g of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4Al + 3O2 —> 2Al2O3
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3
Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2.
Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.
Thus, 1.65 moles of O2 is needed for the reaction.
Finally, we shall determine the mass of O2 needed as shown below:
Mole of O2 = 1.65 moles
Molar mass of O2 = 2 × 16= 32 g/mol
Mass of O2 =?
Mole = mass/Molar mass
1.65 = mass of O2 /32
Cross multiply
Mass of O2 = 1.65 × 32
Mass of O2 = 52.8 g
Therefore, 52.8 g of O2 is needed for the reaction.
Dios mío, vaya, eso es una locura, no lo sabía, pero gracias por la información. :)
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