Answer:
Explanation:
Answer 1:
Lithium : 1s2 2s1 Fluorine: 1s2 2s2 2p5 Carbon: 1s2 2s2 2p2
Argon : 1s2 2s2 2p6 3s2 3p6 Sulphur: 1s2 2s2 2p6 3s2 3p4
Nickel: 1s2 2s2 2p6 3s2 3p6 3d8 4s2 Rubidium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 Xenon: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6
Answer 2: A. Fluorine B. Calcium
C. It is Tellurium if this was the exact electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4 you intend to write, if not, no element has such electonic configuration.
D. Bromine but the correct electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
The molecular formula of sucrose is - C₁₂H₂₂O₁₁
molecular mass of sucrose - 342 g/mol
molarity of sucrose solution is 0.758 M
In 1 L solution the number of sucrose moles are - 0.758 mol
Therefore in 1.55 L solution, sucrose moles are - 0.758 mol/L x 1.55 L
= 1.17 mol
The mass of 1.17 mol of sucrose is - 1.17 mol x 342 g/mol = 4.00 x 10² g
K + I - > KI
Potassium (needs to lose 1 electron) responds with Iodine (needs to pick up 1 electron) to fulfill both component's octet, shaping a salt, potassium iodide
This is a similar case for NaCl, simply unique components. Trust this made a difference.
Answer:
The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
Explanation:
Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.

Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol
Temperature at which Trichlorofluoromethan boils , T= 296.95 K
The molar entropy of the evaporation of Trichlorofluoromethan :

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
Answer:
C.Melt both cubes and look for a broader range of melting temperatures. The one that melts over a broader range of temperatures is the amorphous solid.
Explanation:
Amorphous solids is one that do not have a fixed melting points but melt over a wide range of temperature due to the irregular shape hence its name. Contrariwise crystalline solids, have a fixed and sharp melting point.
This comes in handy to solve the riddle. We can characterise the pair with the melting point property.