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3 years ago

14

Determine the volume of a piece of gold with a mass of 318.97g

1 answer:

djyliett [7]3 years ago

6 0

Answer:

318 g / 19.32 g v = 16. your volume is 16 hope this helps

Explanation:

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Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the s

Annette [7]

Answer:

Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons

Explanation:

The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.

When these electrons change orbit or level they release energy in the form of light, which is known as a photon.

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What does a dissolved salt look like?

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Water but with salt

Explanation:

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3 years ago

A magnet moved near a coil of wire can cause a(n)

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When a magnet moves near a coil of wire it can cause an A. electric current

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4 years ago

Read 2 more answers

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

8 0

3 years ago

La densidad del óxido de magnesio. MgO, es de 3.581 g/cm3 El MgO, es de 3.581 g/cm3 El MgO cristaliza con ordenamiento cúbico co

Jobisdone [24]

Answer:

$a=4.213cm$

$r=1.490x10^{-8}cm$

Explanation:

Hola.

En este caso, para calcular la longitud (a) de una cara de celda unitaria, consideramos la siguiente ecuación:

$\rho =\frac{#at*M}{a^3N_A}$

En la que consideramos el número de átomos por celda (4 para FCC), la masa molar (40.3 g/mol para MgO) y el número de avogadro para obtener:

$3.581g/mol = \frac{4atom/celda*40.3g/mol}{a^3*6.02x10^{23}atom/mol}$

Despejando para a, obtenemos:

$a^3 = \frac{4atom/celda*40.3g/mol}{3.581g/cm^3*6.02x10^{23}atom/mol}\\\\a=\sqrt[3]{7.478cm^3} \\\\a=4.213cm$

Finalmente, el radio lo calculamos como:

$r=\frac{\sqrt{2}*a}{4}=\frac{\sqrt{2}*4.213x10^{-8}cm}{4}\\\\r=1.490x10^{-8}cm$

¡Saludos!

6 0

3 years ago