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IceJOKER [234]
3 years ago
6

Why cant school just be poggers for once:(

Chemistry
2 answers:
pashok25 [27]3 years ago
7 0

Answer:

yes the schools should be poghers for once only:)

Sindrei [870]3 years ago
3 0
I thought I was the only one wondering this but anyways can I have branliest(:
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4
TiliK225 [7]
Given : Density of Bromine = 3.12 g/mL
Formula : Density = Mass / Volume

Part A :
Given Volume = 125 mL
Density = Mass / Volume
So, Mass = Density x Volume
= 3.12 x 125
= 390 grams

Part B :
Given mass = 85.0 gm
Density = Mass / Volume
So, Volume = Mass / Density
= 85 / 3.12
= 27.24 mL
8 0
4 years ago
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Which statement is true about Air at different temperatures
Alchen [17]
Are there any choices?
6 0
4 years ago
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DESPERATE WILL GIVE BRAINLIST AND THANKS FOR COREECT ANSWER
Oksi-84 [34.3K]

Answer:

<h3>the coded blueprint for life </h3>

Explanation:

<h3>I hope it helps ❤❤❤</h3>
8 0
3 years ago
A neutral atom with the electron configuration 2-8-6 would most likely form a bond with an atom having the configuration
Fittoniya [83]

Answer:

The configuration of the atom would be 2-8-2.

Explanation:

Any atom of an element combines with other element to complete its octet and become stable.

The electron configuration of the given atom is 2-8-6. That means the atom has 6 electrons in its outermost shell. To become stable the atom should have 8 electrons in its outermost shell. The given atom has 6 electrons so it either lose 6 electrons or gain 2 electrons to complete its octet.

But we know the atom having 5,6,7 electrons in its outermost shell they do not lose, they gain either 3 or 2 or 1 electrons to complete its octet.

So we say that atom with the electron configuration 2-8-6 bond with the atom having electron configuration 2-8-2.

8 0
4 years ago
A sample of an unknown biochemical compound is found to have a percent composition of 45.46 percent carbon, 7.63 percent hydroge
Leona [35]

Answer:

Formular = C₅H₁₁NO₃

Explanation:

The empirical formular is the simplest formular of a compound can have.

We use the steps below to obtain the empirical formular;

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Carbon = 45.46% = 45.46g

Hydrogen = 7.63% = 7.63g

Nitrogen = 10% = 10g

Oxygen = 100% - (45.46% + 7.63% + 10%) = 36.31% = 36.31g

Step 2: Determine the number of moles of each type of atom present.

Molar amount (M) = m/atomic mass

Carbon = 45.46 / 12 = 3.7883

Hydrogen = 7.63 / 1 = 7.63

Nitrogen = 10 / 14 = 0.7143

Oxygen = 36.91 / 16 = 2.3069

Step 3: Divide the number of moles of each element by the smallest number of moles. Smallest = 0.7143

Carbon = 3.7883 / 0.7143 = 5.3035

Hydrogen = 7.63 / 0.7143 = 10.67

Nitrogen =  0.7143 / 0.7143 = 1

Oxygen = 2.2693 / 0.7143 = 3.1770

Step 4: Convert numbers to whole numbers

Carbon = 5

Hydrogen = 11

Nitrogen = 1

Oxygen = 3

Formular = C₅H₁₁NO₃

3 0
4 years ago
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