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quester [9]
3 years ago
7

Explain why the solutions of SbCl3(aq) in our lab also contain HCl. Why would adding a few drops of this solution to 300 mL of w

ater result in a foggy or cloudy appearance?
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

Here's what I get  

Explanation:

SbCl₃ reacts with water to form slightly soluble antimony oxychloride.

SbCl₃(aq) +H₂O(ℓ) ⇌ SbOCl(s) + 2HCl(aq)

Your observation is an example of Le Châtelier's Principle in action,

The SbCl₃(aq) in your lab has enough HCl added to push the position of equilibrium to the left and keep the SbOCl in solution.

If a few drops of the SbCl₃(aq) were added to 300 mL of water, the solution would turn cloudy. The HCl would be so dilute that the position of equilibrium would lie to the right, and a cloudy precipitate of antimony oxychloride would form.

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HELPPP !!
USPshnik [31]
ANSWER: All of the above
5 0
3 years ago
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Calculate the number of atoms in a 5.31 x 10³ g sample of sodium.<br><br> number of atoms:
Leviafan [203]

Answer:

        no of atoms =   13.9 x 10^25

Explanation:

                No. of moles = mass of compound / molar mass of   compound

   

       As ; mass of sodium = 5.3 x 10^3 g

       

        Molar mass of sodium = 22.9 g/ mol

         

        putting values  

         

         n = 5.3 x 10^3 / 22.9

       

         n = 231.4 mol

Also; no of mol (n) = no of particles / Avagadros number

               so no of particles = n x Avagadros no.

     

               put n = 231.4 and Avagadros no = 6.023 x 10^23

 

                 no of particles = 231.4 x 6.023 x 10^23

                                   =   13.9 x 10^25  

       

             

4 0
2 years ago
Agnesium, calcium, and strontium are examples of ____________.
Ber [7]
D: alkaline earth metals
6 0
3 years ago
The iodide ion concentration in a solution may be determined by the precipitation of lead iodide. Pb2 (aq) 2I-(aq) PbI2(s) A stu
MatroZZZ [7]

Answer:

M_{I^-}=0.6841M

Explanation:

Hello.

In this case, since the precipitation of the lead iodide is related to the iodide ion in solution, if we make react lead (II) nitrate with an iodide-containing salt, a possible chemical reaction would be:

Pb(NO_3)_2+2I^-\rightarrow PbI_2+2NO_3^-

In such a way, since 15.71 mL of a 0.5770-M solution of lead (II) nitrate precipitates out lead (II) iodide, we can first compute the moles of lead (II) nitrate in the solution:

n_{Pb(NO_3)_2}=0.5570\frac{molPb(NO_3)_2}{L}*0.01571L=0.01393molPb(NO_3)_2

Next, since there is a 1:2 mole ratio between lead (II) nitrate and iodide ions, we compute the moles of those ions:

n_{I^-}=0.01393molPb(NO_3)_2*\frac{2molI^-}{1molPb(NO_3)_2} =0.02785molI^-

Finally, since the mixing of the two solutions produce a final volume of 40.71 mL (0.04071 L), the resulting concentration (molarity) of the iodide ions in the student's unknown turns out:

M_{I^-}=\frac{0.02785molI^-}{0.04071L}\\\\ M_{I^-}=0.6841M

Best regards!

5 0
3 years ago
Describe the process used to measure out a specific mass of a solid.
Andre45 [30]
You first weigh you're solid to determine it's mass. Then put it in a container that has liquid with definite volume which can't dissolve solid, absolutely when you put solid in liquid, liquid raise and you have volume of solid. <span><span>
specificmass=mass/volume</span></span>
6 0
3 years ago
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