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Maurinko [17]
3 years ago
10

and the measured absorbance of a sample solution is 0.367, what is the ferroin concentration in the measured solution

Chemistry
1 answer:
Lina20 [59]3 years ago
4 0

This question is incomplete, the complete question is;

If the detailed experimentally determined calibration curve is;

Absorbance = 12,100 M⁻¹ × [ferroin (M)] + 0.00230

and the measured absorbance of a sample solution is 0.367, what is the ferroin concentration in the measured solution?

Choice of Answer

A) 2.76 x 10⁻⁵ M

B) 3.01 x 10⁻⁵ M

C) 3.45 x 10⁻⁵ M

D) none of the above

Answer:

the ferroin concentration in the measured solution is 3.01 × 10⁻⁵M

Option B) 3.01 x 10⁻⁵ M is the correct Answer

Explanation:

Given the data in the question;

Absorbance = 12,100 M⁻¹ × [ferroin (M)] + 0.00230 ------- let this be equation 1

Measured Absorbance = 0.367

so, we simply substitute value of measured absorbance into equation 1

0.367 = 12,100 × [ferroin] + 0.00230

solve for [ferroin ]

0.367 = 12,100  × [ferroin] + 0.00230

12,100 × [ferroin]  = 0.367 - 0.00230

12,100 × [ferroin ]  = 0.3647

[ferroin ]  = 0.3647 / 12,100

[ferroin ]  = 3.014 × 10⁻⁵ ≈ 3.01 × 10⁻⁵M

Therefore,  the ferroin concentration in the measured solution is 3.01 × 10⁻⁵M

Option B) 3.01 x 10⁻⁵ M is the correct Answer

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Answer:

<h2>0.36 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

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From the question we have

n =  \frac{2.17 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{2.17}{6.02}  \\  = 0.360465...

We have the final answer as

<h3>0.36 moles</h3>

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3 years ago
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Answer:

(a) 6.38 × 10⁻²⁹ kg·m·s⁻¹; (b) 7.00 kg·m·s⁻¹; (c) 82.7 µm; (d) 7.53 × 10⁻³⁴  m;

(e) Δx ∝ 1/m

Explanation:

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\begin{array}{rcl}\Delta x \times 0.0700 & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{0.0700}\\\\ &\geq& \textbf{7.53 $\mathbf{\times 10^{-34}}$ m}\\\end{array}

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