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kozerog [31]
3 years ago
12

Iron has a density of 7.86 g/cm3 (1 cm3=1 mL). Calculate the volume (in dL) of a piece of iron having a mass of 3.67 kg . Note t

hat the density is provided in different units of volume and mass than the desired units of volume (dL) and the given units of mass (kg). You will need to express the density in kg/dL (1 cm3 = 1 mL) before calculating the volume for the piece of iron
Chemistry
1 answer:
Liula [17]3 years ago
4 0
<span>7.86 g / 1 mL
divide g by 1000 for kg and divide mL by 100 for dL

</span>0.00786 kg / 0.01 dL = 0.786 kg/dL (density)

3.67 kg / 0.786 dL = 4.6692 dL

Another approach:

3.67kg > (3.67kg x 1000) = 3670 grams 
3670 grams / 7.86 (density) =  466.92 mL
466.92 mL (divide by 100 for dl) = 4.6692 dL
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3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
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Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

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log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

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b)

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= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

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