Answer:
the animal
Explanation:
a rodent thas in the ground
<h3>
Answer:</h3>
2.04 mol CBr₄
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Organic</u>
- Writing Organic Compounds
- Writing Covalent Compounds
- Organic Prefixes
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
675 g CBr₄
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of Br - 79.90 g/mol
Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol
<u>Step 3: Convert</u>
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<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄
Answer is: excess of hydrazine is 16 grams.
Chemical reaction: N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂<span>O(g).
</span>m(N₂H₄) = 80,1 g.
m(N₂O₄) = 92,0 g.
n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).
n(N₂H₄) = 80,1 g ÷ 32 g/mol.
n(N₂H₄) = 2,5 mol.
n(N₂O₄) = 92 g ÷ 92 g/mol.
n(N₂O₄) = 1 mol; limiting reactant.
From chemical reaction: n(N₂H₄) : n(N₂O₄) = 2 : 1.
n(N₂H₄) = 2 mol reacts.
Δn(N₂H₄) = 2,5 mol - 2 mol = 0,5 mol.
Δm(N₂H₄) = 0,5 mol · 32 g/mol = 16 g.
Answer:
Explanation:When silver Ag metal is added to copper sulphate CuSO4 solution, no reaction takes place as silver is less reactive than that of copper and cannot displace copper from its solution. Hence, when a piece of silver metal is added to copper sulphate solution there will be no reaction.