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Lunna [17]
3 years ago
15

The concentration of Fe2 in a sample is determined by measuring the absorbance of its complex with ferroxine. The sample, measur

ed in a 1.00 cm cuvette, has an absorbance of 0.242 . The reagent blank in the same cuvette has an absorbance of 0.041. What would be the absorbance reading for each of these two solutions if measured in a 5.00 cm cuvette
Chemistry
1 answer:
sdas [7]3 years ago
8 0

Answer:

  • Absorbance of sample solution = 1.21
  • Absorbance of reagent blank = 0.205

Explanation:

In order to solve this problem we need to keep in mind the <em>Lambert-Beer law</em>, which states:

  • A = ε*b*C

Where ε is the molar absorption coefficient, b is the length of the cuvette, and C is the concentration.

By looking at the equation above we can see that if ε and C are constant; and b is 5 times higher (5.00 cm vs 1.00 cm) then the absorbance will be 5 times higher as well:

  • Absorbance of sample solution = 0.242 * 5 = 1.21
  • Absorbance of reagent blank = 0.041 * 5 = 0.205
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<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

H_2(g)+F_2(g)\rightarrow 2HF(g)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]

We are given:

\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = \frac{-14.1}{1}\times 2.20=-31.02J/K

Hence, the value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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3 years ago
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(1 point) How many mL of water is added to dilute a 10 mL of 0.4 M solution to a new concentration of 0.1 M?
LiRa [457]

Answer:

The volume of water added is 30 mL

Explanation:

Given data

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We can find V₂ using the dilution rule.

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0.4 M × 10 mL = 0.1 M × V₂

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CH4(g)+ 2 O2(g) + CO2(g) + 2 H20 (l)
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Answer:

<u>∆H° reaction = -890.3 kJ</u>

Explanation:

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Multiply equation of water H2O by 2

and reverse the direction of equation of CH4

Hence the sign of ∆H°c = +74.8 kJ becomes +ve.

We are doing this because CH4 is to be in the reactant side not  in the product side.

∆H° reaction = ∆H°a +2(∆H°b) -∆H°c

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∆H° reaction = -890.3 kJ

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