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1 year ago

a covalent bond involves: a. transferring of b. sharing protons c. transferring protons d. sharing electrons

Chemistry

1 answer:

Bas_tet [7]1 year ago

8 0

A covalent bond involves: d. sharing electrons

The atoms of a molecule are held together by electrical forces that hold the atoms together, this is due to the electrical attraction between the electrons of the atom of one element and the nucleus of the other atom, this force of attraction is called covalent bond.

A covalent bond involves the sharing of electrons that are in the last energy layer of the atoms because they are the most likely or willing to be shared.

An example of a covalent bond is the bond between a hydrogen atom and a carbon atom in methane.

The atom is the smallest part of the composition of matter, it is indivisible and is composed of a nucleus that has protons and neutrons, and around the nucleus there are the electrons.

Learn more about the atom at: brainly.com/question/17545314

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Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96

cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0

2 years ago

What is the pressure of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 300.0 K

Ierofanga [76]

Start with the ideal gas equation, PV=nRT PV=nRT and rearrange for pressure to get p=nRTV p=nRTV . You have all the necessary variables in their proper units, so plug em' into the equation to solve for pressure in units of atmospheres.

P=(2.5 mol)(300 K)(0.08206 L atm K −1 mol −1 50.0 L =1.23 atm P=(2.5 mol)(300 K)(0.08206 L atm K−1 mol−150.0 L=1.23 atm

All that needs to be done now is converting atmospheres to mm Hg Hg .

1.23 atm∗760 mm Hg1 atm =935 mm Hg 1.23 atm∗760 mm Hg1 atm=935 mm Hg .

That value makes sense, since the original pressure in atmospheres was above 1, the pressure in mm Hg Hg will be above 760.

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3 years ago

Who was the 37 persident

sammy [17]

Answer:

Richard Nixon was elected the 37th President of the United States (1969-1974) after previously serving as a U.S. Representative and a U.S. Senator from California.

Explanation:

3 0

2 years ago

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_____ energy is energy available to do work.

Arisa [49]

The answer is C. Free Energy.

Activation energy is the energy that must be exceeded before a chemical reaction can occur. Free energy best describes as energy available to do work.

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2 years ago

Read 2 more answers

Thanks for the help :) Question attached below:

Alekssandra [29.7K]

• Take a look at the steps below to see how to balance this equation. Let's start by writing the unbalanced equation given the information.

Unbalanced Equation : C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O (g) ,

Start by Balancing the Carbons : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + H₂O (g)

Now let's balance the Hydrogen : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + 4H₂O (g)

Balancing the Oxygen : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

Balanced Equation : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

• Let's apply dimensional analysis here,

0.7 L propane × (5 liters Oxygen / 1 liter Propane) = 3.5 Liters of Oxygen

• Similarly we can identify the liters of carbon dioxide produced in the reaction,

0.7 L propane × (3 liters Carbon Dioxide / 1 liter Propane) = 2.1 Liters of Carbon Dioxide

• 0.7 L propane × (4 liter water vapor / 1 liter propane ) = 2.8 Liters of Water Vapor

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3 years ago