First calculate for the molar mass of the given formula unit, CaCO₃. This can be done by adding up the product when the number of atom is multiplied to its individual molar mass as shown below.
molar mass of CaCO₃ = (1 mol Ca)(40 g Ca/mol Ca) + (1 mol C)(12 g of C/1 mol of C) + (3 mols of O)(16 g O/1 mol O) = 100 g/mol of CaCO₃
Then, divide the given amount of substance by the calculated molar mass.
number of moles = (20 g)(1 mol of CaCO₃/100 g)
number of moles = 0.2 moles of CaCO₃
<em>Answer: 0.2 moles</em>
Psolution = X · PH_20
= 0.966 · 31.8 torr
= 30.7 torr
Your answer is D rods and cones. Hope this helps!! :)
It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
Learn more: brainly.com/question/16243729
a. mass of iron = 69.92 g
b. percent yield = 93%
<h3>Further eplanation
</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
a.
Reaction
Fe₂O₃+3CO⇒2Fe+3CO₂
MW Fe₂O₃ : 159.69 g/mol
mol Fe₂O₃
mol Fe₂O₃ : mol Fe = 1 : 2
mol Fe :
mass of Fe(Ar=55.845 g/mol) :
b.
actual yield = 65 g
theoretical yield = 69.92 g
percent yield :
