To solve this problem, we will use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.08205 L·atm·mol-1·K-1)
T = temperature (in K)

Step 1: Convert the volume from mL to L.
1 L = 1000 mL
So, 93 mL = 93/1000 = 0.093 L

Step 2: Convert the temperature from °C to K.
K = °C + 273.15
So, 325 °C = 325 + 273.15 = 598.15 K

Step 3: Substitute the values into the ideal gas law equation and solve for n (number of moles).
2 atm * 0.093 L = n * 0.08205 L·atm·mol-1·K-1 * 598.15 K

Solving for n gives:
n = (2 atm * 0.093 L) / (0.08205 L·atm·mol-1·K-1 * 598.15 K) = 0.0044 mol

So, there are 0.0044 moles of O2(g) present in a 93 mL vessel at a pressure of 2 atm and at 325 °C.

Question

To solve this problem, we will use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.08205 L·atm·mol-1·K-1)
T = temperature (in K)

Step 1: Convert the volume from mL to L.
1 L = 1000 mL
So, 93 mL = 93/1000 = 0.093 L

Step 2: Convert the temperature from °C to K.
K = °C + 273.15
So, 325 °C = 325 + 273.15 = 598.15 K

Step 3: Substitute the values into the ideal gas law equation and solve for n (number of moles).
2 atm * 0.093 L = n * 0.08205 L·atm·mol-1·K-1 * 598.15 K

Solving for n gives:
n = (2 atm * 0.093 L) / (0.08205 L·atm·mol-1·K-1 * 598.15 K) = 0.0044 mol

So, there are 0.0044 moles of O2(g) present in a 93 mL vessel at a pressure of 2 atm and at 325 °C.

Knowee AI · Accepted Answer

To solve this problem, we will use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.08205 L·atm·mol-1·K-1)
T = temperature (in K)

Step 1: Convert the volume from mL to L.
1 L = 1000 mL
So, 93 mL = 93/1000 = 0.093 L

Step 2: Convert the temperature from °C to K.
K = °C + 273.15
So, 325 °C = 325 + 273.15 = 598.15 K

Step 3: Substitute the values into the ideal gas law equation and solve for n (number of moles).
2 atm * 0.093 L = n * 0.08205 L·atm·mol-1·K-1 * 598.15 K

Solving for n gives:
n = (2 atm * 0.093 L) / (0.08205 L·atm·mol-1·K-1 * 598.15 K) = 0.0044 mol

So, there are 0.0044 moles of O2(g) present in a 93 mL vessel at a pressure of 2 atm and at 325 °C.

How many moles of O2(g) are present in a 93 mL vessel at a pressure of 2 atm and at 325 °C? Provide you answer to 4 decimal places.R = 0.08205 L·atm·mol-1·K-1

Question

Solution

Similar Questions

Upgrade your grade with Knowee