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Vsevolod [243]
3 years ago
14

If 250 mL of 0.375 mol/L solution of potassium iodide was mixed with an excess of lead (II) nitrate, what mass of lead (II) iodi

de precipitate would form? Please tell me the steps to solve problem- not just the answer :)
Chemistry
1 answer:
lukranit [14]3 years ago
4 0
The balanced chemical equation is Pb(NO3)2+ 2KI produces PbI2 + 2K(NO)3

.4 L of KI × (.375 mol/L of KI) × (1 mol of PbI2 / 2 mol of KI) × (461 g of PbI2/1 mol of PbI2)

=34. 6 g of PbI2 precipitate.
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The net ionic equation for the reaction of aluminum sulfate and sodium hydroxide
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3 years ago
The CI-B-Cl bond angle in the BCl_3 molecule is O 180° O 90° O 120° 060° 109.5°
noname [10]

Answer:

120°

Explanation:

   Valence electrons of boron = 2

Valence electrons of chlorine = 7  

The total number of the valence electrons  = 3 + 3(7) = 24

The Lewis structure is drawn in such a way that the octet of each atom in the molecule is complete. So,  

The Lewis structure is:

             

               

There is no lone pair involved, so, It is of type AB₃.

According to the theory, the atoms will form a geometry in such a way that there is minimum repulsion and maximum stability.  

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5 0
3 years ago
How much energy is required to move the electron of the hydrogen atom from the 1s to the 2s orbital?
Alja [10]

Answer:

1.63425 × 10^- 18 Joules.

Explanation:

We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.

We will be using the formula in the equation (1) below;

Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).

Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.

Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;

∆E= - R(H) × [1/nf^2 - 1/ni^2 ].

Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.

Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].

= -2.179 × 10^-18 Joules × (0.25 - 1).

= - 2.179 × 10^-18 × (- 0.75).

= 1.63425 × 10^- 18 Joules.

7 0
3 years ago
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