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LenaWriter [7]
3 years ago
12

Post good memes for brainlyest

Chemistry
2 answers:
Nadya [2.5K]3 years ago
8 0

Answer: I don't know why I think this is funny...tbh

Explanation:

Katen [24]3 years ago
7 0

Answer:

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░░░░░▐█▄█▌░░░░░░▀█▄░░

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░░░░▄▄▄██▀▀▀▀░░░░░░░

░░░█▀▄▄▄█░▀▀░░

░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

▄░▐░░░▄▄░█░▀▀ ░░

▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him In Brainly Question,

░░░░░░░▄▄▐▌▄▄░░░ So, He Can Take

░░░░░░░▀███▀█░▄░░ Over Brainly

Explanation:

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Which of the following is not part of our solar system?
andreyandreev [35.5K]

Answer:

none of the above

Explanation:

4 0
3 years ago
Read 2 more answers
Why does celsium have a lower first ionzation energy than rubidiumn.
densk [106]

Answer:

i really have no clue

Explanation:

8 0
2 years ago
Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda
bezimeni [28]

<u>Answer:</u> The average atomic mass of X is 28.09 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

  • <u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

  • <u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]

\text{Average atomic mass of X}=28.09amu

Hence, the average atomic mass of X is 28.09 amu

4 0
3 years ago
Calculate the molecular (formula) mass of each compound: (a) iron(ll) acetate tetrahydrate; (b) sulfur tetrachloride; (c) potass
Nutka1998 [239]

Answer:

a) Iron(ll) acetate tetrahydrate: 245,68 g/mol

b) Sulfur tetrachloride: 173,87 g/mol

c) Potassium ermanganate 158,034 g/mol

Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

The molecule of Iron(II) acetate itetrahydrate is (CH3COO)2Fe•4H2O, it means that you have:

2 atoms of carbon times the atomic weight of C (12.00g/mol)= 24g

14 atoms of Hidrogen times the atomic weight of H (1,00g/mol)= 14g

6 atoms of Oxigen times the atomic weight of O (16,0g/mol)= 96

2 atoms of Iron times the atomic weight of Fe (55,84g/mol)= 111,68g

At last, you only have to add the results: 24+14+96+11,68= 245,68g/mol. This example was for the first molecule.

See you,

5 0
3 years ago
Whose cell walls may contain cellulose can be plantlike, animal like or fungus
lapo4ka [179]

Answer:

I'm not sure

Explanation:

sorry wish that I did tho

8 0
3 years ago
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