Answer: 83.74L
Explanation:
Temp. = 295K
P = 713torr = 0.938atm
Mass = 2.65g
PV = nRT
V = nRT/PV
n = Mass/Molar mass
Molar mass of Hydrogen gas = 1.00784*2= 2.0156g/mok
n = 2.65/2.0156 = 1.31469mol
V = 1.31469*0.08205*295/0.938
V = 83.74L
The volume = 83.74L
Bromine. both Bromine and Mercury are both a liquid at STP. :)
Answer:
In this phenomenon we talk about ideal gases, that is why in these equations the constant is the number of moles and the constant R, which has a value of 0.082
Explanation:
The complete equation would have to be P x V = n x R x T
where n is the number of moles, and if it is not clarified it is because they remain constant, as the question was worded.
On the other hand, the symbol R refers to the ideal gas constant, which declares that a gas behaves like an ideal gas during the reaction, and its value will always be the same, which is why it is called a constant. The value of R = 0.082.
The ideal gas model assumes that the volume of the molecule is zero and the particles do not interact with each other. Most real gases approach this constant within two significant figures, under pressure and temperature conditions sufficiently far from the liquefaction or sublimation point. The real gas equations of state are, in many cases, corrections to the previous one.
The universal constant of ideal gases is not a fundamental constant (therefore, choosing the temperature scale appropriately and using the number of particles, we can have R = 1, although this system of units is not very practical)
Answer:
The partial pressure of argon in the jar is 0.944 kilopascal.
Explanation:
Step 1: Data given
Volume of the jar of air = 25.0 L
Number of moles argon = 0.0104 moles
Temperature = 273 K
Step 2: Calculate the pressure of argon with the ideal gas law
p*V = nRT
p = (nRT)/V
⇒ with n = the number of moles of argon = 0.0104 moles
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 273 K
⇒ with V = the volume of the jar = 25.0 L
p = (0.0104 * 0.0821 * 273)/25.0
p = 0.00932 atm
1 atm =101.3 kPa
0.00932 atm = 101.3 * 0.00932 = 0.944 kPa
The partial pressure of argon in the jar is 0.944 kilopascal.
