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devlian [24]
3 years ago
11

Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma

ny milliliters of 5.00 M HCl ( aq ) are required to react with 3.15 g Zn ( s ) ?
Chemistry
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

We need 19.3 mL of HCl

Explanation:

Step 1: Data given

Molarity HCl = 5.00 M

Mass Zn = 3.15 grams

Step 2: The balanced equation

Zn (s) + 2HCl (aq) ⟶ ZnCl2 (aq) + H2 (g)

Step 3: Calculate moles Zn

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 3.15 grams  / 65.38 g/mol

Moles Zn = 0.0482 moles

Step 4: Calculate moles HCl needed

For 2 moles HCl we need 1 mol Zn to produce 1 mol ZnCl2 and 1 mol H2

For 0.0482 moles Zn we need 2*0.0482 = 0.0964 moles HCl needed

Step 5: Calculate volume needed

Molarity = moles / volumes

Volumes = moles / molarity

Volume HCl needed = 0.0964 moles HCl /5.00 M

Volume HCl needed = 0.01928 L = 19.3 mL

We need 19.3 mL of HCl

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Answer:

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Explanation:

1. 1700 cm⁻¹

A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.

2. Triplet-quartet

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The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.

The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.

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