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3 years ago

Explain why sodium (metal) is soft and can be bend, whereas NaCl is hard and brittle.

Chemistry

1 answer:

julia-pushkina [17]3 years ago

7 0

Answer:

Sodium chloride has a crystalline face-centered cubic structure whereas metallic sodium body-centered cubic structure.

Explanation:

Hello, atomic arrangements provide the molecules different features and behaviors, since the sodium metal has a body-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-body-centered.svg) the lack of inner atoms, allows the material to be soft and bendable. On the other hand the compacted sodium chloride's face-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-face-centered.svg), provides it a crystalline structure which is hard and brittle since the atoms are closer.

Best regards!

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Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

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Answer:

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Explanation:

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Answer:

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A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp

miss Akunina [59]

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

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