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4 years ago

This element has 3 electrons in its Lewis dot structure and is in the 2p orbital section

1 answer:

3 0

Hen two electrons occupy the same orbital, they always have opposite spin. No other knowledge about electron spin is needed for A-level.) Only when the 2s is filled, can the 2p sub-level fill. With the next six elements (boron, carbon, nitrogen, oxygen, fluorine and neon) the three 2p orbitals are filled.

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3 years ago

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4 years ago

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The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

Answer: The experimental van't Hoff factor is 1.21

Explanation:

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

Hence, the experimental van't Hoff factor is 1.21

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3 years ago

f 4.2 moles of copper metal react with 6.3 moles of silver nitrate, how many moles of silver metal can be formed, and how many m

alex41 [277]

1. Balance the reaction
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
2. Find the limiting [divide moles of reactant by their balancing numbers]
Cu: 4.2/1 =4.2
AgNO3: 6.3/2= 3.15
AGNO3 is the limiting
How many moles of silver metal will be formed?
The ratio between the limiting reactant and silver is 2:2 or 1:1
therefore 6.3 moles of silver will be formed.

How many moles of excess reactant will be left?
Since each mole of copper is equal to 2 moles of AgNO3. divide 6.3 by 2 then minus it from coppers moles
6.3/2 = 3.15
4.2-3.15 = 1.05 moles of copper will be left

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3 years ago