Question

Какой объём занимают 1,2*10 в 26-ой степени молекул аммиака (NH3) при нормальных условиях?

Answer 1

V/Vm=N/Na
V=VmN/Na=22,4*1,2*E26/6,01*E23=4472,55 л

Answer 2

Calculating the Volume of Ammonia Molecules

To calculate the volume of 1.2*10^26 ammonia (NH3) molecules at normal conditions, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Given: - Number of ammonia molecules (n) = 1.2*10^26 - Normal conditions typically refer to a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (101.3 kPa).

Using the ideal gas law, we can solve for the volume (V) occupied by the given number of ammonia molecules.

Calculation:

The number of moles (n) can be calculated using Avogadro's number, which is 6.022*10^23 molecules per mole.

n = (1.2*10^26) / (6.022*10^23) = 199.34 moles

Now, we can use the ideal gas law to find the volume (V): V = (nRT) / P Where: - n = 199.34 moles - R = 0.0821 L·atm/(K·mol) (ideal gas constant) - T = 273.15 K (temperature at normal conditions) - P = 1 atm (pressure at normal conditions)

Let's calculate the volume using the ideal gas law.

Result:

The volume occupied by 1.2*10^26 ammonia molecules at normal conditions is approximately 4.92*10^24 liters.