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3 years ago

A compound that has a sea of delocalized electrons has what type of bonding?

Chemistry

1 answer:

Anarel [89]3 years ago

8 0

Answer:

Metallic Bonding

Explanation:

Metallic Bonding

In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

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3 years ago

What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr

Masja [62]

Answer: 116 g of copper

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds = 4.00 hr = $4.00\times 3600s=14400s$ (1hr=3600s)

$Q=24.5A\times 14400s=352800C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500C=193000C$ of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = $\frac{63.5}{193000}\times 352800=116g$ of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus remaining in solution = (0.1-0.003)=0.097moles

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3 years ago

For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of

viktelen [127]

Answer:

$\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%$

Explanation:

Hello!

In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

$n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\$

Now, we compute the atomic percentages as shown below:

$\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%$

Best regards!

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2 years ago