calculate the volume oqupaied by 20gram H2 at NTP

To calculate the volume occupied by 20 grams of H2 at NTP (Normal Temperature and Pressure), we need to use the ideal gas law equation.

The ideal gas law equation is given by:

PV = nRT

Where:
P = pressure (in Pa)
V = volume (in m^3)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in K)

First, let's determine the number of moles of H2. To do this, we need to know the molar mass of H2, which is approximately 2 g/mol (as hydrogen has an atomic mass of 1 g/mol).

Number of moles (n) = mass / molar mass
Number of moles (n) = 20 g / 2 g/mol
Number of moles (n) = 10 mol

Since we are given that the conditions are at NTP, we can set the temperature to 273 K and the pressure to 1 atmosphere (atm), which is approximately equal to 101,325 Pa.

Now we can rearrange the ideal gas law equation to solve for volume:

V = (nRT) / P

V = (10 mol * 8.314 J/(mol·K) * 273 K) / 101325 Pa

V ≈ 0.224 m^3

Therefore, the volume occupied by 20 grams of H2 at NTP is approximately 0.224 cubic meters (m^3).

To calculate the volume occupied by 20 grams of hydrogen gas (H2) at NTP (Normal Temperature and Pressure), you will need to use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

NTP conditions are defined as a pressure of 1 atmosphere and a temperature of 273.15 Kelvin.

First, you will need to convert the mass of hydrogen from grams to moles. The molar mass of hydrogen is 2 grams per mole (H2).

number of moles (n) = mass (m) / molar mass (M)

n = 20 g / 2 g/mol
n = 10 moles

Now you have the number of moles (n), pressure (P), ideal gas constant (R), and temperature (T). You can rearrange the ideal gas law equation to solve for volume (V):

V = nRT / P

V = (10 moles) · (0.0821 L·atm/(mol·K)) · (273.15 K) / 1 atm

V = 224.67 liters

Therefore, the volume occupied by 20 grams of hydrogen gas at NTP is approximately 224.67 liters.