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S_A_V [24]
2 years ago
6

Assume the atomic mass of element x is 22.99 amu. A 22.42 g sample of x combines with 34.56 g of another element y to form a com

pound xy. Calculate the atomic mass of y
Chemistry
2 answers:
zavuch27 [327]2 years ago
6 0
Vhnvhmhvmhvvhm,bmhmnvnvn
ahrayia [7]2 years ago
3 0

Answer:

The atomic mass of y is 35.44 g/mol

Explanation:

Step 1: Data given

The atomic mass of element x is 22.99 amu

A 22.42 g sample of x combines with 34.56 g of another element y to form a compound xy

Step 2: The equation X + Y → XY

Step 3: Calculate moles X

Moles X = mass / atomic mass

Moles X = 22.42 grams / 22.99 amu

Moles X = 0.9752 moles

Step 4: Calculate atomic mass Y

MMy = 34.56 grams / 0.9752 moles

MMY = 35.44 g/mol

The atomic mass of y is 35.44 g/mol

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When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

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The answer should be E,C,F,B,A,D in that order.
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