### Chemistry Lab Preparation TaskA student needs about 200 mL of a 0.8 M MgSO₄ solution to run a reaction in the lab and must make this solution. The student has a vial of solid MgSO₄, a 250.0 ± 0.1 mL volumetric flask, a balance with an error of ± 0.001 g, and deionized water to use in making this solution in the lab.**Hint**: This process is the same as what we did to make the KIO₃ solution in the experiment last week.#### Question:How many significant figures does the final calculated molarity of the MgSO₄ solution have?#### Answer Choices:- 5- 4- 1- 3- 2---**Explanation for Instructors**:- Ensure students remember the concept of significant figures in measurements.- Encourage students to determine the number of significant figures based on the accuracy and precision of the given equipment and measurements.For understanding, let students refer to the significant figures rule:1. Non-zero digits are always significant.2. Any zeros between significant digits are significant.3. Leading zeros are not significant.4. Trailing zeros in a decimal number are significant.Equipment provided and their precisions:- **Volumetric Flask:** 250.0 ± 0.1 mL (4 significant figures in the volume)- **Balance:** ± 0.001 g (3 significant figures in mass measurement)Therefore, the significant figures in the final molarity calculation are determined by the least number of significant figures in any one measurement. Students should use this understanding to answer the question accurately.

Given that 200 mL of a 0.8 M MgSO4 solution.

A student needs about 200 mL of a 0.8 M MgSO4 solution to run a reaction in lab, and must make this solution. The student has a vial of solid MgSO4, a 250.0 ± 0.1 mL volumetric flask, a balance with an error of ± 0.001 g, and deionized water to use in making this solution in the lab. Hint: this process is the same as what we did to make the KIO3 solution in the experiment last week. How many significant figures does the final calculated molarity of the MgSO4 solution have? 5 4 1

A student needs about 200 mL of a 0.8 M MgSO4 solution to run a reaction in lab, and must make this solution. The student has a vial of solid MgSO4, a 250.0 ± 0.1 mL volumetric flask, a balance with an error of ± 0.001 g, and deionized water to use in making this solution in the lab. Hint: this process is the same as what we did to make the KIO3 solution in the experiment last week. How many significant figures does the final calculated molarity of the MgSO4 solution have? 5 4 1

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Indicate how many moles of the second reactant would be required to react exactly with 0.275 mole of the first reactant
. In performing the lab reaction, the student made a mixture of 3 reaction components in a beaker. This was done by mixing 13.00 mL of 0.00250 M Fe³(aq) with 11.00 mL of 0.00250 M SCN (aq) and 6.00 ml. of deionized H₂O. Upon mixing these solutions (total volume of 30.00 mL), the final solution turned a deep. red color. The absorbance of this equilibrium mixture was measured and the concentration of Fe(SCN)2 was determined to be 3.00 x 10 M with the use of the Beer's Law constant (k). Determine the equilibrium constant (K) for the reaction from these values by completing the ICE table below. Be sure to report your final value of K at the bottom. (5 points) 3/15 Initial (mmol) Fe³+ Equilibrium (M) (aq) Fe (aq) 0.008 Change (mmol) -3X105 Equilibrium (mmol)X106 + SCN- (aq) Fet][SCN) 3X105 SCN (aq) = Fe(SCN)2- (aq) Fe(SCN)² (aq) 0 +3×105 X 0.02 3x05 7.7x1556 3 x 105 3.00 x 105 M 2+ Fe ² (aq) + SON (ag) → Fe (SCN) ²+ (aq) (fe+³) = 13X 0.002 = 0.008 30 SCN³) = 30X0.002 0.02 30 Fet ³ (aq) +…
5:15 1 ull 4G Discussion Assignment #6 O o o INFO DISCUSSION Discussion Assignment #6 Due:Wednesday, 14 Oct 2020, 3:00 PM In titration, one solution is added to another solution until a chemical reaction between the components in the solutions has run to completion. The solution being added is called the titrant, and we say that it solution. The completion of the reaction is usually shown by a change of color caused by a substance called an indicator. used titrate the other A typical titration proceeds in the following way. A specific volume of the solution to be titrated is poured into an Erlenmeyer flask. For example, 15.0 mL of a sulfuric acid solution of unknown concentration is added to a 250 mL Erlenmeyer flask. An indicator is then added to the flask, which changes color when the reaction is complete. The titrant in this example is potassium hydroxide, having a concentration of 0.121 M. The titrant is added to a buret. After slowly adding the titrant to the flask, the reaction…
Balance please
Because the limiting reactant in this case will be SCN- (from KSCN), the moles of SCN- in solution will determine the moles (and concentration) of FeSCN²+ in each standard solution. Concentrations and volumes are provided in the experiment procedure in Table 4.1. Use the concentration of KSCN, volume of KSCN and total final volume of each solution (each standard solution will be 20.00 mL) to calculate the FeSCN²+ concentration in each of the four standard solutions. 1 Fe³+ (aq) iron (III) 1. FeSCN²+ concentration in Standard Solution 1 = 234 + 4 SCN- (aq) thiocyanate TABLE 4.1. Volumes for the Part A standard solutions for the Beer's Law Plot. Solution Volume of FeSCN²+ (aq) thiocyanoiron (III) 5.00 5.00 5.00 5.00 Volume of 0.200 M Fe(NO3)3 (ml) 0.00200 M KSCN (mL) 2.00 1.50 1.00 0.50 Reaction 1 Volume of Deionized Water (mL) 13.00 13.50 14.00 14.50 Continued on next page...
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Compare the predictions you made in the preparation activity with your observations of the actual reactions. In which reactions were you able to correctly predict the products? In which reactions did you make incorrect predictions? Were there particular reaction patterns that you had difficulty predicting correctly? Why do you think that was the case?