Calculate the density of oxygen,O2 , under each of the following conditions:
1- STP
2- 1.00 atm and 35.0 C
For part 1, STP is 1.43 which is correct.
1146 joules of heat is released when a 100.0 gram sample of an unknown metal cools from 95.00°C to 65.00°C. Determine the specific heat of a metal.
1.43,1.27
43245
I hate dr bob
To calculate the density of a gas, we need to use the ideal gas law equation and the molar mass of the gas. The ideal gas law equation is given by:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas
The molar mass of oxygen gas (O2) is equal to 32.00 g/mol.
1. STP (Standard Temperature and Pressure):
At STP, the values are:
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/(mol·K)
We need to find the density, which is equal to mass/volume. Since the molar mass is given, we can calculate the mass using the number of moles. Rearranging the ideal gas law equation, we can solve for n:
n = PV / RT
n = (1 atm) × (1 L) / (0.0821 L·atm/(mol·K) × 273.15 K)
n = 0.0408 mol
Next, we calculate the mass (m) using the number of moles (n) and the molar mass (M):
m = n × M
m = 0.0408 mol × 32.00 g/mol
m = 1.31 g
Therefore, the mass of the oxygen gas at STP is 1.31 g. The volume is given as 1 L. Thus, the density (d) is:
d = m / V
d = 1.31 g / 1 L
d = 1.31 g/L
2. 1.00 atm and 35.0°C:
To calculate the density at 1.00 atm and 35.0°C, we need to convert the temperature to Kelvin:
T = 35.0°C + 273.15 = 308.15 K
Using the same formula as above, we can calculate the number of moles (n):
n = PV / RT
n = (1 atm) × (1 L) / (0.0821 L·atm/(mol·K) × 308.15 K)
n = 0.0402 mol
The mass (m) can be calculated as before:
m = n × M
m = 0.0402 mol × 32.00 g/mol
m = 1.29 g
Therefore, the mass of the oxygen gas at 1.00 atm and 35.0°C is 1.29 g. The volume is still given as 1 L. Thus, the density (d) is:
d = m / V
d = 1.29 g / 1 L
d ≈ 1.29 g/L
So, the density of oxygen gas under each of the given conditions is approximately 1.31 g/L at STP and approximately 1.29 g/L at 1.00 atm and 35.0°C.
If its an ideal gas (which it isn't), the gas at STP will occupy 22.4 L/mol.
1 mol = 32 g for oxygen.
density = 32/22.4 = ??
For 308 K, determine how much 22.4 L will occupy from V1/T1 = V2/T2.
Then 32/V2 = ??
Check my thinking. Check my numbers.