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anyanavicka [17]
2 years ago
3

Fill in the table to show your work." Write the entire given; you do not need to include the substance/formula. Round all number

s to thehundredths.Use the following units: "mol", "molecules", "atoms" and "g" for grams and "L" for liters.if you need to write an exponent use the "*" (ex: 6.02 x 10^23).Convert 5.3 x 1028 molecules of co to moles.

Chemistry
1 answer:
Triss [41]2 years ago
8 0

In order to solve this one we have the information that 1 mol is equal to 6.022*10^23 molecules, then we need to find how many moles would be 5.3*10^28 molecules, we can do that by doing the following calculation:

6.022*10^23 = 1 mol

5.3*10^28 = x moles

6.022*10^23x = 5.3*10^28

x = 5.3*10^28/6.022*10^23

x = 88010.63 moles or in scientific notation = 8.80*10^4 moles

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Read 2 more answers
Calculate the solubility product constant, Ksp, of lead(II) chloride, PbCl2, which has a
V125BC [204]

Answer:

0.0159m

Explanation:

9 M

Explanation:

Lead(II) chloride,  

PbCl

2

, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.

Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,  

K

sp

, will be established between the solid lead(II) chloride and the dissolved ions.

PbCl

2(s]

⇌

Pb

2

+

(aq]

+

2

Cl

−

(aq]

Now, the molar solubility of the compound,  

s

, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.

Notice that every mole of lead(II) chloride will produce  

1

mole of lead(II) cations and  

2

moles of chloride anions. Use an ICE table to find the molar solubility of the solid

 

PbCl

2(s]

 

⇌

 

Pb

2

+

(aq]

 

+

 

2

Cl

−

(aq]

I

 

 

 

−

 

 

 

 

 

0

 

 

 

 

 

0

C

 

 

x

−

 

 

 

 

(+s)

 

 

 

 

(

+

2

s

)

E

 

 

x

−

 

 

 

 

 

s

 

 

 

 

 

2

s

By definition, the solubility product constant will be equal to

K

sp

=

[

Pb

2

+

]

⋅

[

Cl

−

]

2

K

sp

=

s

⋅

(

2

s

)

2

=

s

3

This means that the molar solubility of lead(II) chloride will be

4

s

3

=

1.6

⋅

10

−

5

⇒

s

=  √ 1.6 4 ⋅ 10 − 5  = 0.0159 M

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