You need to find the whole molar mass of the compound using the periodic table to add the values.
Na2CO3= (2 x 23.0) + 12.0 + (3 x 16.0)= 106 g/mol
H2O= 10 x [ (2 x 1.01 ) + (16.0) ]= 180.2 g/mol
the total molar mass is 106 + 180.2 = 286.2 g/mol
the percentage of water you can find by doing "parts over the whole"
H2O%= 180.2 / 286.2 X 100= 63.0%
Answer:
A. 4.
Explanation:
<em>no. of moles (n) = mass/molar mass.</em>
mass = 64.0 g, molar mass = 16.0 g/mol.
∴ no. of moles (n) = mass/molar mass = (64.0 g)/(16.0 g/mol) = 4 mol.
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.
Best regards.
.
