102 grams of ammonia is formed when 3 moles of nitrogen and 6.7 moles of hydrogen reacts.
Explanation:
The equation given is of Haeber's process in which the nitrogen is limiting factor in the ammonia formation and hydrogen if in excess gets delimited.
We know that 1 mole of Nitrogen gives 2 moles of ammonia.
We have 3 moles of nitrogen here,
So, 6 moles of ammonia will be form
so from the formula
no of moles=mass/atomic mass
mass= no. of moles*atomic mass
= 6*17
= 102 grams of ammonia will be formed.
So, 6 moles or 102 grams of ammonia is formed when 3 mole of nitrogen and 6.7 mole of hydrogen reacts.
B, they belong to group 1
A: Li is period 2 and Na is period 3
C: Na has one more electron shell
D: Not related to how it is spelled
Answer: Molar concentration of the tree sap have to be 0.783 M
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 19.6 atm
i = Van't hoff factor = 1 (for non-electrolytes)
R = Gas constant =
T = temperature of the solution =
Putting values in above equation, we get:


Thus the molar concentration of the tree sap have to be 0.783 M to achieve this pressure on a day when the temperature is 32°C
1. Convert the number of moles to number of particles (divide 6.41 by 6.022 x 10^23)
2. K2O dissociates into three ions (2 potassium ions and 1 oxide ion), so multiply the result from step 1 by 3 to get your answer