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Natali5045456 [20]
3 years ago
8

Write the number of atoms of each element found in one unit of the compound:

Chemistry
2 answers:
zhenek [66]3 years ago
7 0
A Potassium Iodine : 2 atoms
B Sodium Sulfide : 3 atoms
C Silicon Dioxide : 3 atoms
D Carbonic Acid : 6 atoms
Inessa [10]3 years ago
7 0
A Potassium Iodine : 2 atoms
B Sodium Sulfide : 3 atoms 
C Silicon Dioxide : 3 atoms 
D Carbonic Acid : 6 atoms

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A molecule that forms when atoms of the same element bond together is called a(n)________
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Answer: element.


Justification:


1) A molecule is the union of two or more atoms.


2) When two or more atoms of different elements bond together, the molecule formed is a compound.


For example: one atoms of hydrogen and one atom of chlorine form the molecule HCl, which is a compound (hydrogen chloride).


3) When the molecules is formed by two atoms of the same element, it is not called a compound, but element. Some examples of this are the diatomic gases:


H₂: hydrogen


Cl₂: chlorine


Br₂: bromine


O₂: oxygen


S₂: sulfur


N₂: nitrogen.

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3 years ago
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How do weather models predict the weather?
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Explanation:

<h2><em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h2>
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2 years ago
How does Earth science overlap with life science? A. Earth science involves the study of the composition of matter. B. Earth sci
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3 years ago
Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0
3 years ago
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