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Alex Ar [27]
2 years ago
3

Calculate the volume (in mL) of 6.25 x 10-4 M ferroin solution that needs to be added to a 10.0 mL volumetric flask and diluted

with deionized (DI) water in order to prepare a calibration standard solution with a concentration of 2.50 x 10-5 M ferroin. As part of your preparation for performing this experiment, repeat this calculation for each of the calibration standards you will need to prepare and record the information in your lab notebook so that you have it ready during the lab session. Group of answer choices 0.200 mL 0.400 mL 0.600 mL 0.800 mL none of the above
Chemistry
1 answer:
Alexeev081 [22]2 years ago
8 0

The volume V2 is 0.400 ml

<u>Explanation:</u>

The dilution equation is the product of initial values of molarity and volume which is equal to the product of final values of molarity and volume.

The dilution equation is given by

                          M1 \times V1 = M2 \times V2

where,

M represents the molarity of the solution  

V represents the volume of the solution

M1 and V1 are the initial values  of the solution

M2 and V2 are the final values of the solution

                            M1 \times V1 = M2 \times V2

               (2.5 \times 10^{-5}) \times 10 = (6.25 \times 10^{-4}) \times V2  

                                    V2 = 2.5 \times 10^{-4} / (6.25 \times 10^{-4})  

                                    V2 = 0.4

    The volume V2 is 0.400 ml

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Sulfur is composed of three isotopes: 32S, 33S, and 34S. The atomic masses of these isotopes are given below. 32S: 31.97207 amu
elena-14-01-66 [18.8K]

Answer:

Abundance of 32S is 94.41%

Explanation:

The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:

Average atomic mass = ∑ Atomic mass istope*Abundance

For the sulfur:

32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>

<em>Where X is abundance of 32S and Y abundance of 33S</em>

Also we can write:

1 = X + Y + 0.0422 <em>(2)</em>

0.9578 - X = Y

Because the sum of the abundances = 1

Replacing (2) in (1):

32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422

32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344

-0.9435 = -0.99939X

0.9441  =X

In percentage, abundance of 32S is 94.41%

3 0
1 year ago
Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
Mekhanik [1.2K]

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

                Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

5 0
2 years ago
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A molecule contains 23.24 g iodine (I),
gavmur [86]
Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN
7 0
2 years ago
Identify the conjugate acid base pair <br> H3PO4(ag)+CO32=HCO3-(ag)+HPO42-(ag)
viktelen [127]

Answer:

H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻

Explanation:

An acid is a proton donor; a base is a proton acceptor.

Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.

CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.

The conjugate base is what's left after the acid has given up its proton.

The conjugate acid is what's formed when the base has accepted a proton.

H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.

H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻

acid       base         conj.       conj.

                               base       acid

3 0
2 years ago
By which process is a precipitate most easily separated from the liquid in which it is suspended
Gelneren [198K]
<span>Filtration, if its a precipitate that means its insoluble. </span>
7 0
2 years ago
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