Answer: The volume of tank is 17.5 L.
Explanation:
Given: Mass of methane = 173 g
Pressure = 15.1 atm
Temperature = 298 K
Molar mass of methane is 16.04 g/mol.
Therefore, moles of methane are calculated as follows.
![No. of moles = \frac{mass}{molar mass}\\= \frac{173 g}{16.04 g/mol}\\= 10.78 mol](https://tex.z-dn.net/?f=No.%20of%20moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B173%20g%7D%7B16.04%20g%2Fmol%7D%5C%5C%3D%2010.78%20mol)
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
![PV = nRT\\15.1 atm \times V = 10.78 mol \times 0.0821 L atm/mol K \times 298 K\\V = \frac{10.78 mol \times 0.0821 L atm/mol K \times 298 K}{15.1 atm}\\= 17.5 L](https://tex.z-dn.net/?f=PV%20%3D%20nRT%5C%5C15.1%20atm%20%5Ctimes%20V%20%3D%2010.78%20mol%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20298%20K%5C%5CV%20%3D%20%5Cfrac%7B10.78%20mol%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20298%20K%7D%7B15.1%20atm%7D%5C%5C%3D%2017.5%20L)
Thus, we can conclude that volume of the tank is 17.5 L.
The pressure that will be exerted if four sample of gas are placed in a single 3.5 container is calculated as below
if each gas occupies 675 mmhg
what about 4 gases in the sample
by cross multiplication
= 675 mm hg x 4/1 = 2.7 x10^3mmhg (answer D)
Answer:Calorimetry
Explanation: In this case, an instrument called Calorimeter is used to measure the amount of heat being transferred
Explanation:
H2 (9) + 2 NOg) N20() + H20G)
H2 (M) NO (M) Rate (M*s)
Trial 1 0.30 0.35 2.835 x 10-3
Trial 2 0.60 0.35 1.134 x 10-2
Trial 3 0.60 0.70 2.268 x 10-2
a. What is the order with respect to H2?
Comparing trials 1 and 2,the concentration of H is doubled and that leads to an increase in the rate of the reaction by a factor of 4. This means the order with respect to H is 2.
b. What is the order with respect to NO?
Comparing trials 2 and 3, the concentration of NO is doubled and that leads to an increase in the rate of the reaction by a factor of 2. This means the order with respect to NO is 1.
c. What is the rate equation for this reaction?
rate = k [H]²[NO]
d. Calculate the rate constant for the reaction.
Taking trial 1;
rate = k [H]²[NO]
2.835 x 10-3 = k (0.30)² (0.35)
k = 90 x 10-3 = 0.09 L2 mol-2 s-1
Shape and size has no effect on an objects density