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VMariaS [17]
3 years ago
8

Question 3After creating a Beer's Law plot using standard solutions of Q, you determined the slope of Beer's Law to be 0.515 M-1

. Your unknown solution of Q tested in Part B of the experiment had an absorbance of 0.145. Determine the concentration (in molarity) of the unknown solution Q from Part B.Question 4Refer to the procedure stated in the manual pages for Part A to answer the following question.Using the equation editor embedded in this question, show a sample calculation determining the original concentration of the provided unknown Q in Part A from the diluted concentration calculated in question 3 above.Lab ManualYou have been provided with a solution of unknown Q, the actual molar concentration is listed as the unknown number. Dilute 15.00 mL of the provided solution to a final volume of 50.00 mL. You may only use the equipment and reagents listed above. Be sure to record your unknown number in your notebook. After making your dilution, calculate the concentration of your diluted solution.
Chemistry
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

3. Concentration of the unknown solution = 2.81 *10^-1 M

4. The concentration of diluted solution (Q)  = 0.9385 M

Explanation:

3. Calculating the concentration of the unknown solution Q using the formula, we have;

A = E*e*C

Where;

E = slope = 0.515 M^-1

e =1 cm

A = absorbance = 0.145

C = Concentration of the unknown solution

Substituting into the formula, we have;

A = E*e*C

C = A/E *e

   = 0.145/0.515*1

  = 2.81 *10^-1 M

Therefore, the concentration of unknown solution  = 2.81 *10^-1 M

4. Calculating the concentration of diluted solution, we have;

Let concentration of initial Q = Mi

So, Mi*Vi = Mdiluted * Vdiluted

Mi * 15.00mL = 2.81 *10^-1* 50.00mL

Mi = (2.81 *10^-1* 50.00)/ 15.00

   = 0.9385 M

Therefore, the concentration of diluted solution (Q)  = 0.9385 M

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3 years ago
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3 0
3 years ago
A piece of copper (12.0 g) is heated to 100.0 °C. A piece of chromium (also 12.0 g) is chilled in an ice bath to 0 °C. The speci
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Answer:

(a) Slightly greater than 20.0 °C

(b) T_F=293.46K=20.3^oC

Explanation:

Hello,

In this case, since we are talking about the equilibrium temperature that will be reached when the copper, chromium and water samples get in contact, the following equation is useful to describe such situation:

\Delta H_{water}+\Delta H_{copper}+\Delta H_{chromium}=0\\

Thus, in terms of masses, heat capacities and temperatures we consider the final temperature as the unknown:

m_{water}Cp_{water}(T_F-T_{water})+m_{copper}Cp_{copper}(T_F-T_{copper})+m_{chromium}Cp_{chromium}(T_F-T_{chromium})=0In such a way, by knowing that the heat capacities of copper and chromium are 0.386 and 0.45 J/(g°C) respectively, by solving for the equilibrium temperature one has:

T_F=\frac{m_{water}Cp_{water}T_{water}+m_{Cu}Cp_{Cu}T_{Cu}+m_{Cr}Cp_{Cr}T_{Cr}}{m_{water}Cp_{water}+m_{Cu}Cp_{Cu}+m_{Cr}Cp_{Cr}}

T_F=\frac{200.0g*4.184\frac{J}{g*K}* 293.15K+12.0g*0.386\frac{J}{g*K} *373.15K+12.0g*0.45\frac{J}{g*K}*273.15K}{200.0g*4.184\frac{J}{g*K}+12.0g*0.386\frac{J}{g*K} +12.0g*0.45\frac{J}{g*K}}\\\\T_F=293.46K=20.3^oC

Hence, the resulting temperature of water turns out slightly greater than 20.0 °C.

Best regards.

7 0
3 years ago
How many moles of gas sample are 5.0 L container at 373K and 203kPa
Rom4ik [11]
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p</span>×V=n×R×<span>T

V= </span><span>5.0 L

T= </span><span>373K

p= </span><span>203kPa
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R is </span> universal gas constant, and its value is 8.314 J/mol×<span>K
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Now when we have all necessary date we can calculate the number of moles:

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Oksana_A [137]

Answer:

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Explanation:

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