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morpeh [17]
3 years ago
14

Help meee pleaseeeee

Chemistry
1 answer:
tatiyna3 years ago
7 0

Answer:

see explanation

Explanation:

Using the periodic table, look at the top of each column => These are group numbers ... Typically (for american tables) the numbers are divided into A-Groups and B-Groups... For this post, you need to focus on the A-Groups, or 'Main Group Elements'... Now, the numbers also represent the number of valence (outer shell) electrons in the elements listed below that number. For example, under Group IA, all elements ( H, Li, Na, K, Rb, Cs & F) all have one (1) outer shell electron. All elements under IIA have two outer shell electrons, IIIA, 3 outer shell electrons and so on. The exception is Helium (He) which has only 2 outer shell electrons and is typically listed under Group VIIIA.

So ...

X· => H, Li & Na

X: => He(noble gas exception), Be & Mg

·X: => B & Al

:X: => C & Si

X(5 dots) => N  

X(6 dots) => O

X(7 dots) => F & I

X(8 dots) => Ne

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It is false because electrons have no mass.
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Do all forms of radioactive decay change the identity of the original element?
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Explanation:

<h3>yes, Radioactive decay involves the emission of a particle and/or energy as one atom changes into another. In most instances, the atom changes its identity to become a new element.</h3>

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2 years ago
What kind of bond is the result of the transfer of an electron?
serious [3.7K]

Answer:

Answer choice A

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5 0
3 years ago
Read 2 more answers
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
You want to slow down the following reaction:
zimovet [89]
Iron oxide is rust. So oil would be an inhibitor.
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