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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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sattari [20]

Answer:

Your strategy here will be to use the molar mass of potassium bromide,

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So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

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+

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Use the periodic table to find the molar masses of these two elements. You will find

For K:

M

M

=

39.0963 g mol

−

1

For Br:

M

M

=

79.904 g mol

−

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To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

M

M KBr

=

39.0963 g mol

−

1

+

79.904 g mol

−

1

≈

119 g mol

−

So, if one mole of potassium bromide has a mas of

119 g

m it follows that three moles will have a mass of

3

moles KBr

⋅

molar mass of KBr



119 g

1

mole KBr

=

357 g

You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

360 g

a

a

∣

∣

−−−−−−−−−

Explanation:

<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>

6 0
3 years ago
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BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

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8 0
3 years ago
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Explanation:

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4 0
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