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Lunna [17]
3 years ago
8

Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,00

0. Given that a sample of monodisperse polyacrilonitrile weighs 676.8 g and contains molecules of -(CH2CHCN)n-, calculate n.
Chemistry
1 answer:
pentagon [3]3 years ago
6 0

Answer:

7.68 × 10²⁴

Explanation:

Step 1: Calculate the mass of 1 molecule of the monomer CH₂CHCN

We will get the mass of the monomer by adding the masses of the elements.

mCH₂CHCN = 3 × mC + 3 × mH + 1 × mN

mCH₂CHCN = 3 × 12.01 amu + 3 × 1.01 amu + 1 × 14.01 amu = 53.07 amu

Step 2: Convert the mass of the monomer to grams

We will use the conversion factor 1 amu = 1.66 × 10⁻²⁴ g

53.07 amu × 1.66 × 10⁻²⁴ g/1 amu = 8.81 × 10⁻²³ g

Step 3: Calculate "n"

We will divide the mass of the polymer by the mass of the monomer.

n = 676.8 g / 8.81 × 10⁻²³ g = 7.68 × 10²⁴

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7 0
3 years ago
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C + O2 = CO2
geniusboy [140]

Answer:

44 grams of CO₂ will be formed.

Explanation:

The balanced reaction is:

C + O₂ → CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • C: 1 mole
  • O₂: 1 mole
  • CO₂: 1 mole

Being the molar mass of each compound:

  • C: 12 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole

By stoichiometry the following mass quantities participate in the reaction:

  • C: 1 mole* 12 g/mole= 12 g
  • O₂: 1 mole* 32 g/mole= 32 g
  • CO₂: 1 mole* 44 g/mole= 44 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

If 12 grams of C react, by stoichiometry 32 grams of O₂ react. But you have 40 grams of O₂. Since more mass of O₂ is available than is necessary to react with 12 grams of C, carbon C is the limiting reagent.

Then by stoichiometry of the reaction, you can see that 12 grams of C form 44 grams of CO₂.

<u><em>44 grams of CO₂ will be formed.</em></u>

3 0
3 years ago
Determine the location of the last significant place value by placing a bar over the digit.
Alexeev081 [22]

Answer:

Determine the location of the lost significant place value by placing a bar over the digit.

Explanation:

7 0
3 years ago
PLEASE HELP!!! PLEASE.
adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄:  The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

<u><em>Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl? </em></u>

  • As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
  • The normality of the NaOH and HCl = Their molarity.

<em>∵ (NV)NaOH = (NV)HCl</em>

<em>∴ N of HCl = (NV)NaOH / (V)HC</em>l = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

<u><em>Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?</em></u>

  • As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
  • <em>The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.</em>

<em>∵  (NV)H₂SO₄ = (NV)KOH</em>

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

<em>∴ The concentration of KOH = 7.675 mN = 7.675 mM.</em>

<em></em>

<u><em>Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?</em></u>

  • As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

<em>The normality of NaOH = Molarity of NaOH = 0.5 N.</em>

<em>∵ (NV)H₂SO₄ = (NV)NaOH</em>

<em>∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄</em> = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

<em>∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.</em>

<em></em>

<u><em>Q₄: Explain the difference between an endpoint and equivalence point in a titration.</em></u>

  • The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
  • The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
  • End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
  • The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
6 0
3 years ago
On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and sulfur gives 2.0 L of CO2, 3.0 L of H2O vapor, and 1.0 L of
Murrr4er [49]

Answer:

The empirical formula of the organic compound is  = C_2H_6S_1

Explanation:

At STP, 1 mole of gas occupies 22.4 L of volume.

Moles of CO_2 gas at STP occupying 2.0 L = n

n\times 22.4L=2.0L

n=\frac{2.0 L}{22.4 L}=0.08929 mol

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of H_2O gas at STP occupying 3.0 L = n'

n'\times 22.4L=3.0L

n'=\frac{3.0 L}{22.4 L}=0.1339 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of SO_2 gas at STP occupying 1.0 L = n''

n''\times 22.4L=1.0L

n''=\frac{1.0 L}{22.4 L}=0.04464 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = \frac{0.08920 mol}{0.04464 mol}=2

Hydrogen =  \frac{0.2678 mol}{0.04464 mol}=6

Sulfur = \frac{0.04464 mol}{0.04464 mol}=1

The empirical formula of the organic compound is  = C_2H_6S_1

3 0
3 years ago
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