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3 years ago

How do scientists prevent the nonpolar perfluorocarbons

Chemistry

1 answer:

Pavel [41]3 years ago

3 0

Scientists prevent the nonpolar perfluorocarbons

in Oxygent from separating from the water by adding surfactant.

Explanation:

The collection of oxygen from lungs and transporting it all those tissues present in the human body and taking carbon dioxide back to lungs are the main functions of the hemoglobin. But, instead of this hemoglobin, When perfluorocarbons , a nonpolar chemical is used to perform the same function for the oxygen transportation, the nonpolar chemical may not mix well with the water when this perfluorocarbons is taken through a water-based saline .

For this purpose an addition of a chemical called surfactant is done. This will enhance the chemical bonding between the non polar chemical and water. The perfluorocarbons spreads into some droplets and gets coated with this bonding substance. Thus it gets attached to the water.

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Answer:

O lowering the temperature of the system

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2 years ago

How many molecules are present in the following quantity 0.250 mole of H2O

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3 years ago

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2 years ago

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kifflom [539]

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3 years ago

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The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p

earnstyle [38]

Answer:

The frequency is $f = 2,01 * 10^{15} \ Hz$

Explanation:

From the question we are told that

The energy required to ionize boron is $E_b = 801 KJ/mol$

Generally the ionization energy of boron pre atom is mathematically represented as

$E_a = \frac{E_b}{N_A}$

Here $N_A$ is the Avogadro's constant with value $N_A = 6.022*10^{23}$

$E_a = \frac{801}{6.022*10^{23}}$

=> $E_a = 1.330 *10^{-18} \ J/atom$

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

$E = hf = E_a$

=> $hf = E_a$

Here h is the Planks constant with value $h = 6.626 *10^{-34}$

$f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}$

=> $f = 2,01 * 10^{15} \ Hz$

8 0

2 years ago