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avanturin [10]
4 years ago
11

Would an element with 7 valence electrons be more or less reactive than an element with 3 valence electrons?

Chemistry
1 answer:
Sedbober [7]4 years ago
5 0

Answer:

That would be more

Explanation:

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How are the concentrations of hydrogen ions and hydroxide ions related in an aqueous solution?
Jlenok [28]

Answer:

An Arrhenius acid increases the concentration of hydrogen (H+) ions in an aqueous solution, while an Arrhenius base increases the concentration of hydroxide (OH–) ions in an aqueous solution.

5 0
3 years ago
Read 2 more answers
One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methyl
NikAS [45]

Answer:

Structures are given below.

Explanation:

  • Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
  • H atoms adjacent to Br will be eliminated.
  • 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
  • Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
  • Structure of alkenes are given below.

5 0
3 years ago
aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is
maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0
3 years ago
How many moles of C6Cl6 are in 5.44g
Kazeer [188]

Explanation:

rbrhrhyhggggsdffffffffffv

4 0
3 years ago
Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.
Citrus2011 [14]

Answer:

Explanation:

 The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu  has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so  Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

7 0
3 years ago
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