Answer:
Therefore 500 ml of solution have 0.25 mol of NaCl .. = 14.61 g (ans.)
Explanation:
Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
I don’t think there is another way for you to get an extra switch from your Xbox sorry:((
13% of 0.900 is 0.117
there are 0.117 milliliters of alcohol in a 0.900-L bottle of wine.
Answer:
Non competitive inhibition
Explanation:
Hello,
During enzymatic catalysis, the active sites could be occupied by the very same products' molecules turning out into an inhibition (the reaction starts to slow down since to active places are available for the reagents to react). Nonetheless this inhibition is not competitive as long as the product does not react due to the active sites it is occupying.
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