Answer:
Experimental group
Explanation:
i hopes this helps let me know if im wrong:)
Answer:
39.2 g
Explanation:
- 2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)
First we <u>convert 55.3 grams of Ni₂O₃ into moles of Ni₂O₃</u>, using its<em> molar mass</em>:
- 55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃
Then we <u>convert 0.334 moles of Ni₂O₃ into moles of Ni</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.334 mol Ni₂O₃ *
= 0.668 mol Ni
Finally we <u>calculate how much do 0.668 Ni moles weigh</u>, using the<em> molar mass of Ni </em>:
- 0.668 mol Ni * 58.69 g/mol = 39.2 g
Answer:
43.05 moles of Al needed to react with 28.7 moles of FeO.
Explanation:
Given data:
Moles of FeO = 28.7 mol
Moles of Al needed to react with FeO = ?
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Now we will compare the moles of Al with FeO.
FeO : Al
2 : 3
28.7 : 3/2×28.7 = 43.05 mol
Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.
Answer:
The ΔHrxn for the above equation = 179 kJ/mol
Explanation:
The reaction bond enthalpies are for the reactant;
3 × N-H = 3 × 390 = 1,170 kJ/mol
2 × O=O = 2 × 502 = 1004 kJ/mol
The reaction bond enthalpies are for the product;
3 × N-O = 3 × 201 = 603 kJ/mol
3 × O-H = 3 × 464 = 1,392 kJ/mol
The ΔHrxn for the above equation is therefore;
ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol
This combination in non polar.
