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olasank [31]
3 years ago
8

Which reaction should be used to convert propene into an alkyl halide?

Chemistry
2 answers:
daser333 [38]3 years ago
5 0

Answer:The correct answer is Addition reaction.

Explanation:

Addition reaction; The reaction in which one molecule combines with smaller molecule to give a single. When propene is converted into an alkyl halide addition reaction will take place.

CH_2=CH-CH_3+HX\rightarrow CH_3-CHX-CH_3

Where as in condensation reaction two larger molecule fuse together with release of smaller molecules.

In Elimination reaction in which sum molecules are eliminated from the compounds.

in substitution reaction , is same as of single displacement reaction.

skelet666 [1.2K]3 years ago
4 0
Substitution that is it 
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Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.
Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of acid, H₃PO₄ (Ma) = 0.390 M
  • Volume of acid, H₃PO₄ (Va) = 24.5 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

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brainly.com/question/14356286

5 0
2 years ago
If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give
tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

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  • The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

  • The potential energy at height h = P.E = 6.381\times 10^{-21} J
  • Molar mass of nitrogen gas =  28.0134 g/mol
  • Mass of nitrogen gas molecule = m

m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg

  • The acceleration due to gravity = g = 9.8 m/s^2
  • The maximum height attained by nitrogen gas molecule = h
  • The potential energy is given by:

P.E=mgh

6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

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