Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
Answer:
the air molecules from the cookies diffused faster than the air molecules from the pumpkin
Explanation:
The smell of cookies reaches first because the air molecules from the cookies were able to diffuse faster than the air molecules from the pumpkin.
<em>At constant temperature, the diffusion rates of two molecules would be dependent on the size of the molecules. The bigger the size of the molecule, the more time it will take for the molecule to diffuse and vice versa. </em>
Thus, <u>since the pumpkin and the cookies were baked at the same temperature, the smell of the cookies was perceived first because the air molecules of the cookies are smaller in size than the air molecules of the pumpkin.</u>
Since Methane is assumed an ideal gas, we need to know its moles in each streams.
Therefore, we can use the ideal gas law to find the mole in the container by:
P V=nRT ⇒ n=PV/R T
n=no of moles of the gas = mass/molar mass
Molar mass o f CO2=44g/mol, mass = 44g
P= 25bar = 101000X25Pa=2.5x106Pa
V = 20L = 20dm3 = 0.02m3
T=100C=373K
R=8.314J/mol.K
n1= 2.5x106Pa x 0.02m3 / 8.314J/mol.K x 373K
n1 = 16.1mols
Similarly for stream 2, we have n2 which is
P=1bar = 100000Pa
T= 100C= 373K
V=200L = 0.02m3
n2= 1x105Pa x 0.02m3 / 8.314J/mol.K x 373K
n1 = 0.645mol
So the new stream is an addition of these two streams of methane which has
n3 = n1 + n2 =16.75mols of methane
T=?
V=200L=0.02m3
P = 25Bar = 2.5x106Pa
T= PV/nR
T = 2.5x106Pa x 0.02m3 / 16.75 x 8.314
T=359K
So the final temperature of the gas in the tank is 359K
