Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
Answer:
A inhibitor structure resembles substrate structure B inhibitor binds non covalently at site other than active site.C inhibitors bind covalently and permanently at active site.
Explanation:
A Reversible competitive inhibitors structurally resembles the substrate and competes with the substrate to bind to the active site of the target enzyme.
B Reversible noncompetitive inhibitors binds no covalently at site of the target enzyme that is different from the active site.
C irreversible inhibitors interacts very tightly to the active site of an enzyme by covalent bond which cannot be overcome.
Moving water has kinetic energy
The electron configuration shown is as of Argon.