Answer:
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Explanation:
Answer:
Limiting reactant is NiSO₄
Explanation:
The reaction of aluminum metal with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel is:
2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni
<em>That means 2 moles of Al react with 3 moles of nickel sulfate.</em>
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Moles of Al and NiSO₄ are:
Al: 108g × (1mol / 26.98g) = 4.00 moles of Al
NiSO₄: 464g × (1mol / 154.75g) = 3.00 moles of NiSO₄
For a complete reaction of aluminium there are necessary:
4.00mol Al ₓ ( 3 moles NiSO₄ / 2 moles Al) = 6 moles of NiSO₄
As you have just 3.00 moles of NiSO₄, the <em>limiting reactant is NiSO₄</em>
This represents a primary amine. An amine has a nitrogen group that is connected to three substituents via single bonds. The number of carbon-based substitutents determines whether it is primary, secondary, or tertiary. In this case, since 2 substitutents are just hydrogen atoms, and only one has a carbon-based skeleton, this is a primary amine.
It could be a tide pool but this is pretty vague
Answer:
5.00 mol Mg
10.0 mol Cl
40.0 mol O
Explanation:
Step 1: Given data
Moles of Mg(ClO₄)₂: 5.00 mol
Step 2: Calculate the number of moles of Mg
The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.
5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg
Step 3: Calculate the number of moles of Cl
The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.
5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl
Step 4: Calculate the number of moles of O
The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.
5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O
