Answer:
i. Lead nitrate:
2Pb (NO3)2 Δ= 2PbO+4NO2+O2
ii. Potassium chlorate:
2KClO3 → 2KCl + 3O2↑
iii. clacium carbonate:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
iv. cupric carbonate :
CuCO3 → CuO + CO2↑
Hope this helped
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Answer:
K ^+ , CO3 ^2-
Explanation:
The compound is potassium trioxocarbonate(IV).
It contains cation (potassium ion) and acid radical ( trioxocarbonate (IV) ion).
Since K is in group 1 of the periodic table, it loses one electron to form ion i.e K^1. trioxocarbonate IV ion has a charge of 2-.and so the ions of the compound are as shown in the answer above.
In atom, each proton is a charge of +1
(Answer) 0.166 moles of copper (I) nitride are needed.
Mass of copper (I) oxide = 35.7 g
Moles of copper (I) oxide formed by the reaction = (Mass / molar mass)
= ( 35.7 g / 143.09 g/mol ) = 0.25 moles
According to the balanced chemical equation,
mole ratio of copper (I) nitride and copper (I) oxide = 2: 3
Therefore, moles of copper (I) nitride needed to form 0.25 moles of copper (I) oxide = (2/3 x 0.25) moles = 0.166 moles.
Thus, 0.166 moles of copper (I) nitride are needed to form 0.25 moles of copper (I) oxide.
