Ask question

Ask question

All categories

3 years ago

15

The coinage metals -- copper, silver, and gold -- crystallize in a cubic closest packed structure. Use the density of silver (10

.5 g/cm3) and its molar mass (107.9 g/mol) to calculate an approximate atomic radius for silver.

1 answer:

lara31 [8.8K]3 years ago

8 0

Answer:

The answer to the question is

The approximate atomic radius for silver is 1.445×10⁻⁸ cm

Explanation:

To solve the question we list out the known variables as follows

The density of silver = 10.5 g/cm³

Molar mass of silver = 107.9 g/mol

Number of moles in one cm³ of silver = 10.5/107.9 moles or 0.0973 moles

Avogadro's law states that equal volumes of all substances contain equal number of particles that is one mole of any substance contain 6.022 × 10²³ elementary particles

Therefore one mole of silver contains 6.022 × 10²³ silver atoms

number of moles of silver in a unit cell = 4/6.022 × 10²³ or 6.642 × 10⁻²⁴ mol which has a mass = 6.642 × 10⁻²⁴ mol × molar mass of silver or 7.167× 10⁻²² g

Therefore the volume of a unit cell is given by Volume = mass/density =

7.167× 10⁻²² g/10.5 g/cm³ = 6.83× 10⁻²³ cm³

The diagonal of the face of a unit cell contains four atomic silver radius therefore

That is 4 × silver radius = diagonal of cubic unit cell face

= √2 × ∛(6.83 × 10⁻²³ cm³)

The approximate atomic radius for silver = (1/4) × √2 × ∛(6.83× 10⁻²³ cm³)

= 1.445×10⁻⁸ cm

You might be interested in

Define Acid and alkalis

kicyunya [14]

Answer:

Acids have a pH less than 7. Acids contain lots of hydrogen ions

Alkalis have a pH greater than 7. Alkalis contain lots of hydroxide ions

<u><em></em></u>

<u><em>Please mark as brainliest if answer is right </em></u>

Have a great day, be safe and healthy

Thank u

XD

7 0

2 years ago

Read 2 more answers

Why are only certain colors absorbed by and emitted from an atom?

zavuch27 [327]

The way I would explain it is quite difficult to understand, so this is what Google says. "The wavelength (or equivalently, frequency) of the photon is determined by the difference in energy between the two states. These emitted photons form the element's spectrum. The fact that only certain colors appear in an element's atomic emission spectrum means that only certain frequencies of light are emitted." I hope this helped.

6 0

3 years ago

Photosynthesis produces<br> A.<br> energy<br> B. glucose.<br> c. carbon dioxide.<br> D.<br> water.

marysya [2.9K]

The answer is glucose

4 0

2 years ago

Read 2 more answers

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

How many moles of N2 in 57.1 g of N2?

SpyIntel [72]

We are given –

Mass of $\bf N_2$ is 57.1 g and we are asked to find number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\pink{\bf\longrightarrow { Molar \:mass \:of \: N_2:-} }$

$\qquad$ $\bf \twoheadrightarrow 14\times 2$

$\qquad$ $\bf \twoheadrightarrow 28$

$\qquad$ ____________________

Now,Let's calculate the number of moles present in 57.1 g of $\bf N_2$

$\qquad$ $\purple{\bf\longrightarrow { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}$

$\qquad$ $\bf \twoheadrightarrow \dfrac{57.1}{28}$

$\qquad$ $\bf \twoheadrightarrow 2.04\: moles$

__________________________________

7 0

2 years ago