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ehidna [41]
3 years ago
13

When 2 moles of Na(s) react with H2O(l) to form NaOH(aq) and H2(g) according to the following equation, 369 kJ of energy are evo

lved. 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Is this reaction endothermic or exothermic? What is the value of q? kJ g
Chemistry
1 answer:
enot [183]3 years ago
4 0

Answer:

Exothermic

q = -8.03 kJ/g

Explanation:

<em>When 2 moles of Na(s) react with H₂O(l) to form NaOH(aq) and H₂(g) according to the following equation, 369 kJ of energy are evolved.</em>

The thermochemical equation is:

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)    ΔH = -369 kJ

Since the energy is evolved, the reaction is exothermic, which is why the enthalpy of the reaction (ΔH) is negative.

The heat released (q) per gram of Na(s) is (molar mass 22.98 g/mol):

\frac{-369kJ}{2molNa} .\frac{1molNa}{22.98gNa} =-8.03kJ/g

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Answer:

Hello attached below is the data found in Aleks Data tab

answer :

i) N0

ii) N0

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Explanation:

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ii) For MgCl2

solubility does not change with pH hence the answer = NO

iii) For Ba(OH) 2

Solubility does change with pH hence the answer = YES

and the pH at which the highest solubility will occur is = 5

attached below is the reason for the answers given

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The spontaneous intermingling of atoms among different substances is accomplished by_____. active transport energy combustion di
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Calculate the solubility product constant, Ksp, of lead(II) chloride, PbCl2, which has a
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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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