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sleet_krkn [62]
3 years ago
13

The most common form of nylon is 63.68% carbon, 9.80% hydrogen, 12.38% nitrogen and 14.14 % oxygen. Calculate the empirical form

ula for nylon.
Chemistry
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

The empirical formule is C6H11NO

Explanation:

Step 1: Data given

Suppose the mass = 100 grams

Nylon contains:

63.68 % carbon = 63.68 grams

9.80 % hydrogen = 9.80 grams

12.38 % nitrogen = 12.38 grams

14.14 % oxygen = 14.14 grams  

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass nitrogen = 14.01 g/mol

Molar mass oxygen = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / moles

Moles C = 63.68 grams / 12.01 g/mol

Moles C = 5.302 moles

Moles H = 9.80 grams / 1.01 g/mol

Moles H = 9.70 moles

Moles N = 12.38 grams / 14.01 g/mol

Moles N = 0.8837 moles

Moles O = 14.14 grams / 16.00 g/mol

Moles O = 0.8838 moles

Step 3: Calculate mol ratio

We divide by the smalleste amount of moles

C: 5.302/0.8837 = 6

H: 9.70/0.8837 = 11

N: 0.8837/0.8837 = 1

O: 0.8838/0.8837 = 1

The empirical formule is C6H11NO

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Question 23 (3 points)
Mice21 [21]

Answer:

<h2>The answer is 3.0 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass of aluminum = 8.1 g

density = 2.7 g/mL

It's volume is

volume =  \frac{8.1}{2.7}  \\

We have the final answer as

<h3>3.0 mL</h3>

Hope this helps you

3 0
3 years ago
Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?
STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

  • For a weak acid like HF, the dissociation of HF will be:

<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>

[H₃O⁺] = [F⁻].

<em>∵ [H₃O⁺] = √Ka.C,</em>

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>

<em></em>

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>

6 0
3 years ago
In the spring of 1984, concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecul
Lady bird [3.3K]

869.6 × 10¹⁴ molecules of EDB

Explanation:

We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.

We need to transform lb in grams.

1 lb = 453.6 grams

1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams

Now we determine the number of molecules of EDB in the sample by devise the following reasoning:

if we have        31.5 × 10⁻⁹ g of EDB in 1 g of sample

then we have   X  g of EDB in 861.8 g of sample

X = (31.5 × 10⁻⁹ × 861.8) / 1 = 27146.7 × 10⁻⁹ g of EDB

Molecular mass of EDB (C₂H₄Br₂) = 188 g/mole

Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:

if       188 g of EDB contains 6.022 × 10²³ molecules

then 27146.7 × 10⁻⁹ g of EDB contains Y molecules

Y = (27146.7 × 10⁻⁹ × 6.022 × 10²³) / 188 = 869.6 × 10¹⁴ molecules of EDB

Learn more:

about Avogadro’s number

brainly.com/question/1445383

#learnwithBrainly

8 0
3 years ago
Read 2 more answers
Which of the following questions can be answered by science? (2 points) What makes a song sound beautiful? What is the meaning o
34kurt

Answer:

D

Explanation:

Its the only answer that actually makes sense and i got it right on the quiz.

5 0
3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

4 0
3 years ago
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