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3 years ago

1. Find the number of protons, neutrons, electrons and electronic structure of flourine 919F.

Chemistry

1 answer:

77julia77 [94]3 years ago

4 0

1. ₉¹⁹F

2. Mg(OH)₂

3. 2H₂ + O₂ ⇒ 2H₂O

<h3>Further explanation</h3>

1. Fluorine, atomic number : 9 , mass number = 19

Symbol : ₉¹⁹F

protons=electrons=atomic number = 9

neutrons = mass number - atomic mass

$\tt n=19-9=10$

Configuration : [He] 2s² 2p⁵

2. Magnesium hydroxide is an ionic compound and is a strong base consisting of 2 ions:

Positive ion: Magnesium: Mg²⁺

negative ion: Hydroxide: OH⁻

The charges of the two are crossed, so that the compound becomes:

Mg(OH)₂

3. Reaction :

H₂ + O₂ --------> H₂O

give coefficient :

aH₂ + bO₂ --------> H₂O

H, left = 2a, right 2⇒2a=2⇒a=1

O, left = 2b, right 1⇒2b=1⇒b=0.5

Reaction becomes :

H₂ + 0.5O₂ --------> H₂O x 2

2H₂ + O₂ --------> 2H₂O

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A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes

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At the end of mitosis, 2n = 20

At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

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A gas containing 80% CH4 and 20% He is sent through a quart/ diffusion tube (see Figure P8.8) to recover the helium, Twenty perc

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For a gas containing 80% CH4 and 20% He is sent through a quart diffusion tube, the composition is mathematically given as

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<h3>What is the composition of the waste gas if 100 kg moles of gas are processed per minute?</h3>

Generally, the equation for the Material balance is mathematically given as

F=R+W

Therefore

100=0.20*1000+W

W=80kmol/min

In conclusion, waste gas compose

2.0/100*100=50/100*20+%*80

Hence

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%CH4=87.5%

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