Answer:
(a) The final pressure of the sample becomes one-fourth of the original pressure.
(b) The pressure of the sample remains unchanged.
(c) The final pressure of the sample becomes four times of the original pressure.
Explanation:
(a)

The volume of sample doubled and kelvin temperature halved.



Therefore, the final pressure of the sample becomes one-fourth of the original pressure.
(b)
Volume and temperature of sample doubled.



Therefore, the pressure of the sample unchanged.
(c)
Volume of sample halved and temperature double.



Therefore, the pressure of the sample becomes four times of the original pressure.
Explanation:
E maybe i am not sure or C
There were two methods that Lord Rayleigh used in his experiment. First, nitrogen was obtained from the atmosphere by passing air over hot copper. Second, air was bubbled through an ammonia mixture, then passed over hot copper. The second method would have a greater mass of nitrogen than the first, <em>because the nitrogen from the atmosphere was contaminated by the nitrogen from the ammonia mixture.</em>
The density of the product gas mixture is 5.39 g/L
Explanation:
<u>Given:</u>
Temperature after the reaction =
C
Pressure in the container = 1.65 atm
The ideal gas equation PV = nRT
<u>To find:</u> Density of the product gas mixture
<u>Step 1:</u>
Molarity (M) = P/(RT)
![\[M=\frac{P}{R T}\]$M=\frac{1.65 \mathrm{atm}}{\left(0.08206 \frac{L \cdot a t m}{m o l . K}\right)(298.15 K)}$](https://tex.z-dn.net/?f=%5C%5BM%3D%5Cfrac%7BP%7D%7BR%20T%7D%5C%5D%24M%3D%5Cfrac%7B1.65%20%5Cmathrm%7Batm%7D%7D%7B%5Cleft%280.08206%20%5Cfrac%7BL%20%5Ccdot%20a%20t%20m%7D%7Bm%20o%20l%20.%20K%7D%5Cright%29%28298.15%20K%29%7D%24)
M = 0.0674 mol/L
<u>Step 2: </u>
The product of the given reaction is
. Its molar mass is
1S x 32.066 g/mol = 32.066 g/mol
3O x 16 g/mol = 48 g/mol
Adding both we get 80.066 g/mol
Therefore the density of the product gas mixture is 
= 5.39 g/L