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Pepsi [2]
3 years ago
13

NF3 has three F atoms bonded to a central N atom, with one lone electron pair on the N. What is the shape of this molecule?

Chemistry
2 answers:
Tems11 [23]3 years ago
6 0

Answer: Trigonal Pyramidal

Explanation: Apex

Black_prince [1.1K]3 years ago
5 0
 <span>When you write the Lewis structure you see that you have three bonding pairs and one lone pair on nitrogen, the central atom. 

What (a) is asking for is what is called the "electron pair geometry", which in your case is tetrahedral. As the name implies, the electron pair geometry looks only at the electron pairs, and not whether they are bonding or not. The electron pair geometry will establish what geometry the molecule can be. 

Question (b) is asking for the "molecular geometry", which is trigonal pyramidal. "Trigonal pyramidal" is withing the "tetrahedral" family of electron pair geometries. The molecular geometry looks only at the shape created by the atoms which are bonded to the central atom. The N and 3 F's form a trigonal pyramid, not a tetrahedron, since there is not a fourth atom bonding at the lone pair. </span>
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
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Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

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<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

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W_{solvent} = Mass of solvent (water) = ? g

Putting values in above equation, we get:

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