The electronic configuration of a chlorine ion in BeCl2 compound is
[2.8.8]^- (answer B)
chlorine atom gain on electron form Be to form chloride ions
chlorine atom has a electronic configuration of 2.8.7 and it gains one electron to form chloride ion with 2.8.8 electronic configuration
FLOOR. Most of the molecules that dont sick to something fall on the floor so most would be on the FLOOR.
<u>Answer: </u>The volume of the solution is 85.7 mL
<u>Explanation:</u>
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(1)
We are given:
Molarity of solution = 0.600 M
Given mass of 
= 12.00 g
We know, molar mass of ![BaSO_4=[(1\times 137.33)+(1\times 32.07)+(4\times 16)]=233.4g/mol](https://tex.z-dn.net/?f=BaSO_4%3D%5B%281%5Ctimes%20137.33%29%2B%281%5Ctimes%2032.07%29%2B%284%5Ctimes%2016%29%5D%3D233.4g%2Fmol)
Putting values in equation 1, we get:
The rule of significant number that is applied for the problems having multiplication and division:
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
Here, the least number of significant figures is 3 that is determined by the number, 0.600. Thus, the answer must have these many significant figures only.
Hence, the volume of the solution is 85.7 mL
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
If you wanted to find a sample of fermium,which has an atomic mass of 100 I would look deep under the earth
