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AveGali [126]
4 years ago
5

Astudent prepareda calibration curve by plotting absorbance of the standards against the [FeSCN2+] molar concentration (M). The

best fit line of the linear equation for the student's experiment was createdas y= 0.1827x-0.0513 (Spec 20). Determine the equilibrium concentration of iron in unknown solution if the absorbance for the unknown solution was measured to be 0.308.
Chemistry
1 answer:
Nuetrik [128]4 years ago
6 0

Answer:

Explanation:

Chemistry 1B Experiment 7

1-3 5.0 1.5 3.5

Part 2: Determining the equilibrium constant.

Label 5 medium-sized test tubes. Table 7.2 shows the amounts of 2.00 × 10–3

M

Fe(NO3)3 (in 1 M HNO3) solution, 2.00 × 10–3

M KSCN solution, and purified water

that should be added to each tube. Pipet the approximate amount of each solution into

each tube. (Record the exact amount of each solution that you actually add. You will

need to use these actual amounts in your calculations.)

Obtain five separate small pieces of parafilm. Close the top of each test tube with

the parafilm. Mix each solution thoroughly by inverting the test tube several times.

Record your observations.

Measure and record the absorbance of each solution at the 447 nm.

Table 7.2 Composition of solutions for determining the equilibrium constant.

Test Tube

Volume of

2.00 × 10–3

M Fe(NO3)3

in 1 M HNO3 (mL)

Volume of

2.00 × 10–3

M KSCN

(mL)

Volume of

purified water

(mL)

2-1 5.0 1.0 4.0

2-2 5.0 2.0 3.0

2-3 5.0 3.0 2.0

2-4 5.0 4.0 1.0

2-5 5.0 5.0 none

Calculations

Part 1. Graphing the relationship between absorbance and [FeSCN2+].

Assuming that “all” of the SCN–

ions have been converted to FeSCN2+ ions,

calculate [FeSCN2+] in each of the solutions in Part 1. For example, in test tube 1-2, 1.0

mL of a 2.00 × 10–3

M KSCN solution was diluted to 10.0 mL. The concentration of

SCN–

that results from this dilution is the one to use for determining [FeSCN2+].

Because of the 1:1 stoichiometry, that initial concentration of SCN– is equal to

[FeSCN2+].

Plot a full-page graph of the absorbance against the concentration of FeSCN2+ in

all standard solutions. Use a ruler to draw the best straight line that comes closest to each

of your five data points. Your line should pass through (0 M, 0). (Why?) This graph is

your calibration curve. When you measure the absorbance of a solution that contains an

unknown concentration of FeSCN2+ ions, you can use this calibration curve to determine

the unknown concentration

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Why LED is better option for testing the conduction of electricity?
Anettt [7]

Answer:

Led is a light-emitting diode, a semiconductor diode that glows when a voltage is applied. So, there is always a sure result of whether a substance conducts electricity or not as the light glows. That is why it is a better option for testing conduction of electricity.

Explanation:

5 0
3 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
Potassium hydroxide is classified as an arrhenius base because koh contains
mixas84 [53]

Answer: hydroxide ions

Explanation:

According to the Arrhenius concept, an acid is a substance that ionizes in the water to give hydronium ion or hydrogen ion  and a bases is a substance that ionizes in the water to give hydroxide ion .

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

As KOH can give hydroxide ions on dissociation , it is considered as arrhenius base.

KOH\rightarrow K^++OH^-

8 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
A car is traveling at 50mph. How long will it take the car to travel 300 miles?
scoundrel [369]

Answer:

6 hours

Explanation:

If it is traveling at 50mph and needs to reah 300 miles, to figure this problem out you would want to divide 300 by 50 to get 6 hours. So it will take six hours until the car travels 300 miles

Hope this helps.

3 0
3 years ago
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