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Alla [95]
4 years ago
14

David found that water can be created in a lab by burning hydrogen gas in air. He concluded that water is not a compound because

only hydrogen was used to form water. What is wrong with David's conclusion
Chemistry
1 answer:
Virty [35]4 years ago
7 0

Hi~~

H2O, commonly known as water, is definitely a compound.

David's conclusion is wrong because the hydrogen gas combined with oxygen from the air to form dihydrogen monoxide, also known as water.


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Why is it important to perform the reagent tests last?
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3 years ago
. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m
stealth61 [152]

Answer:

<h2>         162g/mol</h2>

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

<em>Nicotine has the formula   </em>C_xH_yN_z<em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>

  • <em>1.0 mol of CO₂</em>
  • <em>0.70 mol of H₂O</em>
  • <em>0.20 mol of NO₂</em>

<em>Assume that all the atoms in nicotine are present as products </em>

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

  • 1.0 mol of CO₂ has 1.0 mol of C
  • 0.70 mol of H₂O has 1.4 mol of H
  • 0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

  • C: 1.0
  • H: 1.4
  • N: 0.20

Divide all by the smallest number: 0.20

  • C: 1.0 / 0.20 = 5
  • H: 1.4 / 0.20 = 7
  • N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

  • C:  5mol  × 12g/mol = 60g
  • H: 7mol × 1g/mol = 7 g
  • N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

5 0
4 years ago
An octapeptide composed of four repeating glycylalanyl units has one free amino group on a glycyl residue and one free carboxyl
SCORPION-xisa [38]

Answer:

Following are the solution to this question:

Explanation:

The octapeptides are: gly-ala-gly-ala-gly-ala-gly-ala

Its structure includes free glycine amino, and alanine residues free carboxylic. Its peptides are customarily extracted from the left side free NH2 residues as well as the right side free -COOH residues only in the same direction, the sequence is provided.

please find the structure in the attached file.

5 0
3 years ago
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