1 mol of any particles has 6.02 * 10 ²³ particles.
If we look at 1 NH3 (1 mol NH3 or 1 molecule NH3), we can see that 1 molecule NH3 has 1 atom of N and 3 atoms of H; also 1 mole of NH3 has 1 mole of N atoms and 3 moles of H atoms.
So, 1 mol of NH3 has 1 mol of N atoms,
and 2.79 mol NH3 have 2.79 mol of N atoms.
2.79 mol of N atoms* 6.02 * 10 ²³ N atoms/ 1 mol N atoms = 1.68*10²⁴ N-atoms
Answer is 1.68*10²⁴ N-atoms.
Answer:
km/h
mph
iph
Explanation:
I have no idea what mis or dit could be
km/h is kilometers per hour
mph is miles per hour
I assume iph is inches per hour?
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
Answer:
The choice of the answer is fourth option that is -61 degrees.
Therefore the temperature drop is -61°Centigrade.
Explanation:
Given:
The temperature in a town started out at 55 degrees
Start temperature = 55°Centigrade. (Initial temperature)
End of the Day = -6°Centigrade. (Final temperature)
To Find:
How far did the temperature drop?
Solution:
We will have,
Substituting the above values in it we get
Therefore the temperature drop is -61°Centigrade.
