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galben [10]
3 years ago
10

The energy needed to remove an electron from an atom is called __________ energy.

Chemistry
2 answers:
MariettaO [177]3 years ago
8 0
<span>The energy needed to remove an electron from an atom is called ionization energy.</span>
aev [14]3 years ago
3 0

Answer:  ionization

Explanation:

Ejection of electrons occur when incident light of frequency greater than threshold frequency falls on a metal, the electrons are ejected from metal.

Ionization is the amount of energy requires to remove the most loosely bound electron from an isolated gaseous atom

Valence is the outermost shell which contains the most loosely bound electron.

Electronegativity is the tendency of an atom to attract the shared pair of electrons towards itself.

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The answer is A

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Because the x would be smalle than |-40|

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Fission and Fusion of Atomic Nuclei:
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C B A D

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Is honey homogeneous or heterogeneous
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Honey<span> is a </span>homogeneous<span> mixture because it has the properties that define</span>homogeneous<span> solutions or mixtures.</span>
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3 years ago
A substance has a volume of 10.0 cm³ and a mass of 89 grams. What is its density?
lara [203]

Answer:

8.9 g / cm^3

Explanation:

Density = mass / volume

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2 years ago
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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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