First we have to write balanced chemical equation:
CuSO₄.5H₂O + 4 NH₃ → [Cu(NH₃)₄]SO₄.H₂O + 4 H₂O
then calculate the molar mass of both CuSO₄.5H₂O and [Cu(NH₃)₄]SO₄.H₂O
MM CuSO₄.5H₂O = 249.612 g/mole
MM [Cu(NH₃)₄]SO₄.H₂O = 245.612 g/mole
from equation we see that:
1 mole CuSO₄.5H₂O ---gives ---> 1 mole [Cu(NH₃)₄]SO₄.H₂O
249.612 g ---gives ---> 245.612 g
? grams ---gives ---> 6 g complex
By cross multiplication we get that:
mass of CuSO₄.5H₂O required = 6.1 g
Ask your mum she smarter than u
Answer:
nitrate, carbonate, chloride, bromide, cyanide, and sulfate.
Explanation:
Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.
