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Radda [10]
3 years ago
14

I need help with this question!!

Chemistry
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

Fossil fuels..........

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The number of protons in an atom is known as its .......... ...........?​
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Which hydrocarbon name adheres correctly to the IUPAC naming system? 3–ethyl–2,4–dimethylhexane 3–methethyl–2,dimethylhexane 4–d
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4 0
3 years ago
What is the empirical formula of a vanadium (i) oxide given that 20.38 grams of vanadium combines with oxygen to form 23.58 gram
oksano4ka [1.4K]

V₂O is the empirical formula of a vanadium (i) oxide given that 20.38 grams of vanadium combines with oxygen to form 23.58 grams of the oxide.

The simplest whole number ratio of the atoms of the elements in a specific compound is shown by the empirical formula for that compound. When the mass of each component of a compound or its % composition by mass is known, an empirical formula may typically be computed.

% composition is equal to (mass of an element minus mass of the compound) / 100%.

Given

Vanadium weighs 20.38 g.

The oxide's mass is 23.58 g.

% Vanadium composition: (20.38 g/23.58 g) 100%

= 86.43%

Oxygen mass equals 23.58 g minus 20.38 g

= 3.2 g

100% of oxygen's composition is (3.2/g/23.58 g)

= 13.57%

Atomic mass and number are related by composition percentage.

Vanadium's atomic weight is 50.94 g/mol.

Atomic weight of V = 86.43 / 50.94

= 1.6967

Oxygen has an atomic mass of 16 g/mol.

O atom count is 13.57/16.

= 0.8481

Now on calculation the ratio of Vanadium to the oxygen is

1.6967/0.8481 = 2/1

Ratio of whole numbers is 2:1

Thus V₂O is the empirical formula .

Learn more about empirical formula here brainly.com/question/13058832

#SPJ4

6 0
1 year ago
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