Answer:
dunno :'(
Step-by-step explanations:
Answer is: 89 g of CaCO₃ is <span>equired to prepare 50.0g of CaO.
</span>Chemical reaction: CaCO₃ → CaO + CO₂.
m(CaO) = 50,0 g.
m(CaCO₃) = ?
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 50,0 g ÷ 56 g/mol.
n(CaO) = 0,89 mol.
from reaction: n(CaCO₃) : n(CaO) = 1 : 1.
n(CaCO₃) = n(CaO).
n(CaCO₃) = 0,89 mol.
m(CaCO₃) = m(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 0,89 mol · 100 g/mol
m(CaCO₃) = 89 g.
n - amount of substance.
I believe the answer is A
<span>You have the answer and you don’t even know it. Just take the inverse of what one electron weights in grams (9.11E-28) to get the number of electrons in one gram. You should get an astonishing large number (1.10E+27). Just in case you do not recognize the notation it is scientific notation as a calculator would write it.</span>