1) Ca-37, with a half-life of 181.1(10) ms.
The answer is D. Fertilizer and vinegar
Answer:
Explanation:
2Al + 3Cl2 --> 2AlCl3
Ag2O + Be -----> 2Ag + BeO
4K + O2 ---> 2K2O
CoP + 3NaBr ---> CoBr3 + Na3P
Si + Hg2S ---> SiS2 + Hg
6Li + Ni3N2 --- > 2Li3N + 3Ni
3Ca + 2K3N --> Ca3N2 + 6K
SnS2 + 2F2 --> SnF4 + 2S
Answer is: <span>1.0 mol X left over.
</span>Chemical reaction: X + 2Y → XY₂.
n(X) = 3,0 mol, excess reactant.
n(Y) = 4,0 mol, limiting reagent.
n - amount of substance.
from reaction: n(X) : n(Y) = 1 : 2.
n(X) : 4 mol = 1 : 2.
n(X) = 2 mol, that reacts.
excess of X: 3 mol - 2 mol = 1 mol.
