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Yakvenalex [24]
3 years ago
13

1.33*10^ 23 molecules of carbon dioxide to moles

Chemistry
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

0.22moles of carbon dioxide

Explanation:

Given parameters:

Number of molecules of carbon dioxide  = 1.33 x 10²³molecules

Unknown:

Number of moles   = ?

Solution:

To solve this problem, we use the relationship below:

   1 mole of a substance contains 6.02 x 10²³moleccules;

 

So;

    1.33 x 10²³molecules will contain \frac{1.33 x 10^{23} }{6.02 x 10^{23} }   = 0.22moles of carbon dioxide

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What is the freezing point of an aqueous solution that has 25.00 g of calcium iodide dissolved in 1250 g of water?
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Answer:

<u></u>

  • <u>- 0.380ºC</u>

Explanation:

The lowering of the freezing point of a solvent is a colligative property ruled by the formula:

  • \Delta T_f=K_f\times m\times i

Where:

  • ΔTf is the lowering of the freezing point
  • Kf is the molal freezing constant of the solvent: 1.86 °C/m
  • m is the molality of the solution
  • i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.

<u />

<u>a) molality, m</u>

  • m = number of moles of solute/ kg of solvent
  • number of moles of CaI₂ = mass in grams/ molar mass
  • number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
  • m = 0.0850667mol/1.25 kg = 0.068053m

<u>b) i</u>

  • Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3

<u />

<u>c) Freezing point lowering</u>

  • ΔTf =  1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC

<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>
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3 years ago
Which two of Earth's spheres interact when the rainfall in a region increases after a volcanic eruption?
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Answer:

Geosphere and Hydrosphere

Explanation:

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Aluminum finds application as a foil for wrapping food stuves. Why?
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15. Which compounds are reactants in the process of cellular respiration?
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Explanation:

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3 years ago
The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0
3 years ago
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