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NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is 4.880 A.(a) Calculate the ionic radius of $\mathrm{H}^{-}$ . (The ionic radius of $\mathrm{Li}^{+}$ is 0.0 .95 $\mathrm{A} . )$(b) Calculate the density of NaH.

a) 1.73 $\mathrm{A}$b) 2.86 $\mathrm{gcm}^{-3}$

02:48

Aadit S.

Chemistry 102

Chapter 10

Liquids and Solids

Liquids

Solids

Carleton College

University of Kentucky

Brown University

University of Toronto

Lectures

04:08

In physics, a solid is a s…

03:07

A liquid is a nearly incom…

02:02

Use the $2.16-\mathrm{g} /…

01:32

A drawing of the NaCl unit…

07:34

Potassium chloride has the…

01:22

Thallium(I) iodide crystal…

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At room temperature and pr…

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The CsCl structure is a si…

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Potassium bromide has the …

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The spacing of adjacent at…

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MnO has either the NaCl ty…

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$\textbf{Density of NaCl}$…

01:23

Sodium has a density of 0.…

04:43

A unit cell of cesium chlo…

03:36

(II) Common salt, NaCl, ha…

03:51

01:40

Barium crystallizes in a b…

02:17

23.8 Cesium chloride, brom…

02:41

NaCl has a lattice energy …

01:45

The crystal structure of s…

04:22

02:05

in this podcast, we're continuing to look at the lattice structures. More specifically will be looking a sodium hydride sodium chloride. So the first thing we're looking for is the radius off the hydride. I am. So the equation that we take is Pythagoras is there? Room D squared is equal to a squared at a squad, so we get D equal to root a square at a squared. Why we get d is able to route to a substitute in our value. 4.80 Armstrong D is equal to root two times 4.880 Armstrong What d is equal to 6.90 Armstrong. So now we let the r plus be the radius of the sodium Ah, minus B the radius of the hydride I am so d is equal to r minus. Add to our minus at AR minus, so d people to fall Art minus Where are minors is equal to D, divided by four substitute in our valley, that is 6.90 I'm strong. So four R plus is equal to 6.9 They were Armstrong, while minus is equal to 6.90 Armstrong divided by four r is equal to 1.73 armstrong. So the ionic radius of the hydride iron is 1.73 Armstrong moving on to the second part here. So we have the massive one mole of sodium being 23 g and the mass of one mole of hydride iron being 1 g. The number of atoms in one mole of sodium or hydrogen is six point. There were 22 times, 10 to the 23 that fall well, and a H unit cell has a mass of 9.66 times tense negative 23 g. One on a H unit. So my way, considering the hydride irons is 15.277 times tense. Negative 20 3 g. Add those together for the total 15.94 time center negative, 23 g. So our value we know fee is equal to a cube. A is Thea edge length, so B is equal to 4.88 zero I'm strong. Cube B is 116.2 times 10 to the negative 24 next week and take a look at the density, so the density is the mass by by the volume, so take a fresh page. So we have 15 0.94 times 10 to the negative. 23 g Maths divided by the volume that we just calculated. 116.2 times tense. Negative. 24 centimeters. Cute. We get a density that is represented by that strange P off 2.86 g percent to me. Too cute. All right.

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