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rjkz [21]
2 years ago
8

A HOMOGENEOUS LIQUID THAT CANNOT BE SEPARATED INTO ITS COMPONENTS BY DISTILLATION BUT CAN BE DECOMPOSED BY ELECTROLYSIS IS CLASS

IFIED AS A/AN _______________.
ELEMENT
SUBSTANCE
COMPOUND
SOLUTION
Chemistry
1 answer:
nata0808 [166]2 years ago
7 0

Answer:

ELEMENTS

Explanation:

CUZ AN A

ELEMENT IS A GROUP OF ATOMS THAT CANNOT BE BROKEN DOWN BY ANY CHEMICAL OR PHYSICAL MEAN

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What is the electron configuration for<br> 08<br> 16
Elis [28]

The electron configuration for Oxygen : [He] 2s²2p⁴

<h3>Further explanation</h3>

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d, and f. The maximum number of electrons for each sub-shell is  

• s: 2 electrons  

• p: 6 electrons  

• d: 10 electrons and  

• f: 14 electrons  

Charging electrons in the sub-shell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

The element is Oxgen, with symbol O, and :

the atomic number=8=number of electron

the atomic mass=16

The electron configuration based on the number of electrons(for Oxygen=8), so the configuration :

\tt _8^{16}O:1s^22s^22p^4 or we can write with noble gas [He] 2s²2p⁴

3 0
2 years ago
How many valence electrons does chlorine have?
Advocard [28]
Valence Electrons has \boxed{7} chlorine
3 0
3 years ago
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. How many moles of ammonia gas, NH3, are required to fill a volume of 50 liters at STP?
murzikaleks [220]
In STP every 22.4 litters is 1 mol
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3 years ago
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Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio
Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

 50.48 % O = 50.48 grams

 49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0
3 years ago
In this photograph what rock formation is presented by the blue lines?
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It is representing syncline rock formation

there are two rock formation based on fold formation : syncline and anticline

In syncline rock formation there is fold like trough unlike anticline where it is like crust

In syncline the fold is downward as shown in photo and the new rock is outer fold and old at inner side

3 0
3 years ago
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