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2 years ago

Which type of bond is formed between atoms of solid gold

Chemistry

2 answers:

goldfiish [28.3K]2 years ago

5 0

Answer:

covalent bonds

Explanation:

maxonik [38]2 years ago

5 0

The type of bond formed is known as a metallic bond

In order to identify the type of bond formed between atoms, you must know the requirements needed for each:

A covalent bond can only have non-metals. In this type of bonding, electrons are shared between atoms.

An ionic bond has both non-metals and metals. In this type of bonding, electrons are transferred. In other words, one atom gives electrons to another atom.

Finally, a metallic bond has only metals. In this type of bonding, electrons are delocalized. Essentially, it means that electrons just float around the metal atoms, and aren't designated to a specific atom. (This quality is the reason why metals are so conductive)

If you have any more questions, don't be afraid to ask!

Good luck :))

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3 years ago

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3 years ago

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2 years ago

Read 2 more answers

Consider the following reaction where Kc = 1.80×10-2 at 698 K:2 HI (g) H2 (g) + I2 (g)A reaction mixture was found to contain 0.

Iteru [2.4K]

Answer:

Answer for the given statements: (1) T , (2) F , (3) T , (4) F , (5) F

Explanation:

At the given interval, concentration of HI = $\frac{0.304}{1}M=0.304M$

Concentration of $H_{2}$ = $\frac{5.07\times 10^{-2}}{1}M=5.07\times 10^{-2}M$

Concentration of $I_{2}$ = $\frac{4.57\times 10^{-2}}{1}M=4.57\times 10^{-2}M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[H_{2}][I_{2}]}{[HI]^{2}}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(5.07\times 10^{-2})\times (4.57\times 10^{-2})}{(0.304)^{2}}=2.51\times 10^{-2}$

So, the reaction is not at equilibrium.

As $Q_{c}> K_{c}$ therefore reaction must run in reverse direction to reduce $Q_{c}$ and make it equal to $K_{c}$ . That means HI(g) must be produced and $H_{2}$ must be consumed.

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2 years ago

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Andre45 [30]

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