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3 years ago

Which property does a metal with a large number of free-flowing electrons most likely have?

Chemistry

2 answers:

Akimi4 [234]3 years ago

6 0

a metal with a large number of free-flowing electrons most likely have good conductivity

astra-53 [7]3 years ago

3 0

Answer: Option (d) is the correct answer.

Explanation:

Ability of an element or substance to conduct electricity is known as conductivity. Also, flow of electric current is the flow of free electrons.

It is known that metals contain more number of free electrons and they are good conductors of heat and electricity. Metals are hard substances.

Whereas non-metals are brittle and they are poor conductors of heat and electricity.

Thus, we can conclude that the property of a metal with a large number of free-flowing electrons most likely have good conductivity.

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What is a polyprotic buffer?

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3 years ago

A metal foil has a threshold frequency of 5.45 \times 10^{14} \text{ } \text{Hz}5.45Ã—10 14 Hz. Which of the colors of visible l

nalin [4]

Answer:

Green, Blue, Indigo, Violet

Explanation:

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3 years ago

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How many moles of atoms are in 9.00 g of 13c?

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which structures are found in eukaryotic cells a.chloroplast b. membrane bound organelles c. mitochondria d. nucleus e. pili

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3 years ago

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0

3 years ago