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Zanzabum
3 years ago
12

Choose a likely identity for X, Y, and Z in these structures. A central X atom is bonded to two chlorine atoms through single bo

nds. There are three lone pairs of electrons on each chlorine atom. C Be B F O N A central Y atom is bonded to three chlorine atoms through single bonds. There are three lone pairs of electrons on each chlorine atom. A central Z atom is bonded to two chlorine atoms through single bonds. There are three lone pairs of electrons on each chlorine atom and two lone pairs of electrons on the Z atom. Be C N B F O C Be B O N F
Chemistry
1 answer:
11111nata11111 [884]3 years ago
8 0

Answer:

X= Be

Y= B

Z=O

Explanation:

From the description of the compound XCl2, among the options listed only beryllium can form such compound with three lone pairs in the two chlorine atoms and no lone pair on the central atom X.

From the description of YCl3, only Boron among the options listed can form such a compound with no lone pair on the central atom and three lone pairs on each of the chlorine atoms.

From the description of ZCl2, only oxygen forms the compound OCl2 among the elements listed where oxygen possesses two lone pairs and each chlorine atom possesses three lone pairs each.

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
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Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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The distillation flask and liebig condenser are usually located above the round bottom flask in the set up.

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