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stealth61 [152]
3 years ago
5

Lead is malleable, so it can be pounded into flat sheets without breaking. How does the bonding within lead help to explain this

property?
A. Metallic bonds involve valence electrons that are removed from one atom and given to another, so the pounding helps the electrons move.

B. Covalent bonds involve valence electrons that are shared between two metal atoms, so the bonds are strong enough to resist the pounding.

C. Metallic bonds involve many valence electrons shared by many atoms, so the bonds can move around as the metal is pounded.

D. Covalent bonds involve valence electrons that are removed from one atom and given to another, so the pounding helps the electrons move.

PLSSS HURRYYY
Chemistry
1 answer:
damaskus [11]3 years ago
8 0

Answer:

correct answer would be c

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Feliz [49]

Answer:

D has to be based on facts so d would be the answer and idea that can only be proven true those are facts not hypothesis or guesses

Explanation:

based on a body of facts that have been repeatedly confirmed through observation and experiment. Such fact-supported theories are not "guesses" but reliable accounts of the real world."

6 0
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Help me!!!!!!!<br><br> A. 2<br> B. 3<br> C.4<br> D.6
riadik2000 [5.3K]

Answer:

A. 2

Explanation:

You need to balance the other side so the reactants and products are equal.

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3 years ago
As distance between two objects increases, gravitational force:
just olya [345]

B.) Gravitational force decreases.

You were right !!

3 0
4 years ago
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What is the molecular mass
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100.08 is the molecular mass.
5 0
2 years ago
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
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