For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer:
Allows diffusion of sub...(true)
its found in both anim...(true)
produces en...(false)
Answer:
Explanation:
Hello there!
In this case, since the reaction for the formation of ammonia is:
We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:
And the result is:
Best regards!
Given the molar mass of Nitrogen is 14.01g/mol you can use that to solve for the moles of nitrogen.
0.235g(1mol/14.01g) = .0168 moles.
A sample of a compound contains 60.0 g C and 5.05 g H.
divide by molar mass of C(12) and H(1) to get molar ratio
C: 60/12=5 and H: 5/1=5
so C:H=5:5=1:1
total molar mass=78
divide by 1C+1H to find the formula: 78/(12+1)=78/13=6
compound is C6H6
