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3 years ago

Freezing point of carbon dioxide a.meter b.liter c.kilogram d.kelvin

1 answer:

7 0

The Freezing point is ALWAYS the same as the melting point,even for different elements. 6,422 is its melting,and freezing point. Same with gold,iron,hydrogen,and many others

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What do atoms when they share electron pairs

ra1l [238]

Covalent bonds I think

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3 years ago

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Which discipline of Earth and space science is concerned with the study of the planets and stars?

sweet-ann [11.9K]

Answer:

The discipline of Earth and space science is concerned with the study of the planets and stars is

A. Astronomy

Explanation:

A.Astronomy:

'Astron' means stars and 'nomos' means laws. So astronomy is the study of stars , planets and space

B. biology

It is derived from Greek word : 'Bios' means life and 'logos' means study.

Biology is study of life, living organism ,structure of organisms,evolution

C.Geology

It is the earth science which includes the study of rocks and solid earth, what it is made of and how it has changed .

D. Oceanography

It is the study of physical and biological aspects of oceans

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3 years ago

Please help me, I am confused on this.

weeeeeb [17]

Answer:

The First choice is correct

Explanation:

That is the closest example of what is shown

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3 years ago

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RSB [31]

The answer is the second choice.

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3 years ago

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What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

3 years ago