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2 years ago

Which electrons are lost by an atom of iron when it forms the Fe3+ ion?

2 answers:

8 0

The two s Orbital electrons and one d orbital electron, are the electrons that are lost by an atom of Iron when it forms the Fe3 + ion.

Varvara68 [4.7K]2 years ago

5 0

B. Two s orbital electrons and one d orbital electron

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450.0 mL of an unknown gas has its pressure decreased from 0.07 atm to 0.04 atm. Assuming temperature remains constant, what wou

tino4ka555 [31]

The final volume of the unknown gas is 787.5 ml

Given:

volume of unknown gas = 450.0 mL

initial pressure of unknown gas = 0.07 atm

final pressure of unknown gas = 0.04 atm

To Find:

final volume of the unknown gas

Solution:

Substituting these values into Boyle’s law, we get

P1V1 = P2V2

(0.07)(450) = (0.04)V2

V2 = (0.07)(450)/(0.04)

V2 = 787.5

So, final volume of the unknown gas is 787.5 ml

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1 year ago

John Dalton thought that atoms

zloy xaker [14]

Answer:

cannot be broken down further

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1 year ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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2 years ago

Compared to carbon dioxide, oxygen has a relatively _____ solubility coefficient and so requires a _____ partial pressure gradie

Andrej [43]

Oxygen has a relatively low solubility coefficient and therefore requires a steep (high) partial pressure gradient to help diffuse the gas into the blood.

Solubility is described as the limiting amount of an element that can dissolve in any amount of solvent at a set temperature. Since oxygen has a low coefficient of this, it requires the help of a higher partial pressure gradient to diffuse properly into the bloodstream.

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2 years ago

How water deposits soil sediment and rock

Oliga [24]

Water deposits soil ,sediments ,and rock by moving them to a different place .Say like a beach .The ocean is currently moving the bits of sand and (glass:sand)rock pieces and always ending up in new places.Same as a river.!!!!

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2 years ago