The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.
To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.
From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.
1) From pV = nRT, n = pV / RT
Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K
n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol
mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042
=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol
Now from a periodic table or a table you get that the molar mass of He is 4g/mol
So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
Answer: Option (A) is the correct answer.
Explanation:
Braising means first of all fry a dish slightly and then cook it slowly in a closed vessel or dish. The vessel is close so that the liquid present inside it does not evaporates.
Also, Braising is done to mix the flavors of different liquids or spices appropriately.
Thus, we can conclude that as a cooking method, braising is valued for its ability to retain flavor.
Answer:
Option d. 2615.0g
Explanation:
Let M1, M2, and M3 represent the masses of the three different samples
M1 = 0.1568934 kg = 156.8934g
M2 = 1.215mg = 1.215x10^-3 = 0.001215g
M3 = 2458.1g
Total Mass = M1 + M2 + M3
Total Mass = 156.8934 + 0.001215 + 2458.1
Total Mass = 2614.994651g
Total Mass = 2615.0g
I think it’s 7.41 because you count up all the atoms and find out how many are x (the large grey ones) and you do 2/27 x 100 which gives you 7.41 :) (sorry if i counted wrong it’s kinda hard)
Answer:
At -13 
, the gas would occupy 1.30L at 210.0 kPa.
Explanation:
Let's assume the gas behaves ideally.
As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-
where 
and 
are initial and final pressure respectively.
and 
are initial and final volume respectively.
and 
are initial and final temperature in kelvin scale respectively.
Here 
, 
, 
, 
and 
Hence 
So at -13 , the gas would occupy 1.30L at 210.0 kPa.
