CHM 2046 Lecture Notes - Lecture 12: Strontium Hydroxide, Decimal Mark, Electronegativity

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azureclam319

11 Jul 2019

School

University of Florida

Department

Chemistry

Course

CHM 2046

Professor

Maria V Korolev

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Document Summary

**anything with a ph/poh/pka/panything you don"t count sig figs as anything before the decimal point. Because the m is not a p value, you do normal sig figs. 0. 15-x x x x^2/(0. 15-x)=1. 8x10^-5 you can ignore the x in the denominator x^2/0. 15=1. 8x10^-5 x=0. 00164316 m h+ and log(h+)=ph! **two sigfigs after decimal because it is a ph. Hcl is a strong acid so it completely dissociates and mol ratios are all 0. 15!! so h also is 0. 15 and log(0. 15) is 0. 82. Hcn is weak and becomes h+ and cn-, X^2/. 005-x=6. 2e-10 and we ignore bottom x and solve (. 005)(6. 2e-10) under the square root that number divided by og concentration (. 005) and multiply by 100 to get % **0. 05m to 0. 005m goes from 0. 011% and 0. 035% means that being more diluted for a weak acid means it can dissociate more! So the next strongest acid or lowest pka is hf, formic, then phenol.

Related Questions

1. Determine the amount of solute in the following samples:

500 ml of 0.2 M KI (potassium iodide)

220 ml of 0.03 M H2SO4 (sulfuric acid)

2.5 l of 0.1 M MgNO3 (magnesium nitrate)

1 ml of 5 M HCl (hydrochloric acid)

This problem requires that you know what molarity (M) is. M (molarity) = # of moles of solutes / Liter of solution. You are given the volume. You should be able to obtain moles of solutes. Your answers can be in moles or grams (either one will be acceptable).

2. How many ml of 0.1 M HCl would be required to titrate 35 ml of 1 M NaOH to the end point?

This is an acid/base reaction.

1) First, write out the balanced chemical reaction for HCl + NaOH.

2) Second, figure out the number of moles of NaOH in 35 ml of 1 M NaOH.

3) How many moles of HCl is required for completely reacting with the moles of NaOH in Part 2).

3. What is the pH of a 0.03 M solution of the strong base NaOH?

1) This question requires that you know pH = - log [H₃O⁺], or simply as pH = - log [H⁺] (to make it easier).

2)So, you have to know how to use your scientific calculator to do log function (check Youtube).

3) This question also requires that you know: pOH = - log [OH^-].

4) This question also requires that you know: [H₃O⁺][OH^-] = 1.0 x 10^-14 or pH + pOH = 14

5) First step, determine the concentration of OH^- in 0.03M NaOH when it dissociates or breaks up into Na⁺ and OH^-.

6) This will give you [OH^-]. You can plug this into pOH = - log [OH^-] to get pOH.

7) Once you get pOH, then use (pH + pOH = 14) to get pH.

4. Balance the chemical reactions shown below and then write equilibrium expressions for each. All reactants and all products are in the gas phase.

CO + F₂ ------> COF₂

CH₄ + O₂ ------> CO₂ + H₂O

N₂O₄ -----> Br₂ + NO

BrNO -----> Br₂ + NO

1)This problem requires that you know how to balance chemical equations.

2)After you balance the equation, you need to know the equilibrium expression

5. Determine the oxidation states of each element comprising the following ionic compounds:

KClO₄

NaMnO₄

Sr(NO₃)₂

NaHCO₃

CH₄

122 BEYOND THE BACK OF THE ENVELOPE 9 is substituted into Eq. 7, a nonlinear equation The equation could, in principle, be solved eaton fot ( before, we will look for an approximate analytical solutacly, but me that the 5 Ã 10'"M of sulfur are en that H,SO, is completely dissociated to HSO;H completely dissociated to SOH. Then for eve three types of sulfate added, there are two mole of all moles of H' in solution or H1-2x 10 10 (10) Is this consistent with Eqs. 7 and 9? If Eq 10 is substituted Eq. 9, we calculate that H,Sol -10 110 11 not ather very mu a pt of 5.6. V iged in the cal EXERCISE 2 10-13 4 If Eqs. 10 and 11 are substituted into Eq. 7, 10' 10-134 101.4 10-134 10-14 . (12) od for every 2n So, indeed, Eq. 10 is correct and the pH of the precipitationis EXERCISE ted globally w fomly in glob pH=-logo [H+)-4. (13) This value is actually a good round-number estimate of the average of precipitation in regions of the world (such as northern Europe and the northeastern United States) where acid precipitation is a se s problem. You may wonder what happens to the rest of the SO. me of the SO, forms sulfate that is deposited in rain or snow in Canada, on the Atlantic Ocean, or in the southeastern United States. Much of the rest falls to earth in the form of dry particulate matter in a process called "dry deposition." This dry sulfate can form sulfuric articles fon be? pH EXERC niou So the last para way of a sh one of the acid when rain water eventually dissolves it. In an elegant pair ot Oppenheimer (1983a, 1983b) has applied box-model concepts to the analysis of regional acid precipitation formation a midwestern United States will result in a p H deposition in rain and snow in that area. way, it is illuminating go nd dispersal. His analysis shows that a reduction in SO, emissions in the eastern and onal reduction in Now that we have worked out the answer the long and rigorous g to go back and look at a shortcut. In paragraph Iscussion, above, we estimated the molar concentration tion to be 5 x 10-5 moles(Syliter. When H,SO, dis-

Decomposition of Malachite

Here is what Malachite is Made Out Of

Malachite is a famous and very popular semi-precious stone. Its banded light and dark green bands are one-of-a-kind, and give it a unique ornamental quality unlike that of any other stone. The light and dark green bands are so distinctive that malachite may be one of the most easily recognized minerals by the general public. Malachite is popular in jewelry, especially Native American southwestern jewelry. The stones inlayed in silver make a nice alternative to traditional turquoise jewelry.

Malachite is a form of copper carbonate with the molecular formula Cu₂CO₃(OH)₂. When heated to around 200 ï¿½C, malachite decomposes into copper (II) oxide (CuO) and other products.

Reaction Stoichiometry

The Law of Conservation of Mass states that matter can neither be created nor destroyed. Therefore, the number of atoms of each element on the reactants side must be same as the number of atoms on the products side. This quantitative study of chemical reactions is called stoichiometry. The substances in a balanced reaction are related to each other by stoichiometric molar ratios. For example, magnesium reacts with hydrochloric acid in a 1:2 stoichiometric ratio in the balanced equation below.

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

The molar ratios can be used as conversion factors to find the amount (in moles) of one substance given the amount of another. For example, given the number of moles of HCl, the amount of MgCl₂ can be determined as shown below.

nMgCl2=nHCl x (1mol MgCl2) / (2mol HCL)

The molar ratios are used to determine the amount in moles. However, the amount of reactant or product is often measured in grams. Mass can be converted to moles and vice versa as shown below.

n = m / (MM)

m = n x MM

where n is the number of moles, m is the mass of the product in g, and MM is its molar mass in g/mol.

To observe the Law of Conservation of Mass, you can perform different types of chemical reactions. For example, you can use a decomposition reaction to observe one compound being broken down into two or more elements or new compounds. For this reaction to take place, you must break existing bonds within the starting compound. Different sources of energy such as heat, light, or electricity may be used. Through a decomposition reaction, you can observe that the atoms in the original compound are not destroyed but are rearranged to form simpler substances, such as elements or compounds.

About This Lab

In this lab, you will test if the relationship between malachite and copper (II) oxide is stoichiometric. To do this, you will heat a known mass of malachite to produce copper (II) oxide. You will then compare the amount of copper (II) oxide obtained experimentally with the amount of copper (II) oxide expected from the balanced equation for the decomposition reaction.

1.Take a crucible from the Containers shelf and place it on the workbench.

2.Take a balance from the Instruments shelf and place it on the workbench.

3.Move the crucible onto the balance. Record the mass of the empty crucible to the nearest 0.001g in your Lab Notes.

4.Move the crucible onto the workbench.

5.Take copper (II) carbonate hydroxide from the Materials shelf and add 10 g to the crucible.

6.Move the crucible onto the balance and record the mass of the crucible containing copper (II) carbonate hydroxide in your Lab Notes.

7.Take a Bunsen burner from the Instruments shelf and place it on the workbench.

8.Move the crucible onto the Bunsen burner.

9.Double-click on the crucible to open its properties. Select Show Contents and click OK. This gives you a cut-away view of the crucible's contents.

10.Turn the Bunsen burner on to low flame by clicking the knob at the bottom once. Clicking multiple times will increase the intensity of the flame until it is turned off.

11.Watch the crucible until the malachite fully decomposes. You will see a distinct color change.

12.Move the crucible onto the workbench to cool.

13.Turn off your Bunsen burner.

14.Weigh the crucible with its contents and record the mass to the nearest 0.001 g in your Lab Notes.

Clear your station.

Lab Notes:

Numbers below refer to numbers in the above âAbout this Labâ

3. Empty crucible = 88 g

5. Added 10 g of Copper (II) carbonate hydroxide to crucible

6. Crucible with Copper (II) carbonate hydroxide = 98 g

11. to Black

14. after heating crucible with its contents = 95.195

Questions I have answered:

How many grams of malachite did you add to the crucible? 10g

How many grams of product were in the crucible after heating? 7.195

By how much did the mass of the crucible's contents change after heating? -2.805

The molar mass of malachite is 221.1 g/mol. How many moles of malachite were present before the reaction? 0.04523 mol

The product remaining in the crucible after heating is copper (II) oxide. The molar mass of copper (II) oxide is 79.55 g/mol. How many moles of copper (II) oxide were produced during the reaction?

0.0945 mol

Please answer the following:

Q1 a. Which of the following represents the balanced chemical equation for this reaction?

Cu2CO3(OH)2=2CuO+H2O

Cu2CO3(OH)2=CuO+CO2

Cu2CO3(OH)2=2CuO+H2O+CO2

Cu2CO3(OH)2=CuO+H2O+CO2

b. Based on the number of moles of malachite that you started with, how many grams of water were produced? The molar mass of water is 18.0153 g/mol. Choose the closest answer.

1.815 g 0.8148 g 1.630 g 1.982 g

c. Based on the number of moles of malachite that you started with, how many moles of CO₂were produced? Choose the closest answer.

1.000 mol 0.04523 mol 0.9000 mol 0.09045 mol

d. Based on the number of moles of malachite that you started with, how many grams of CO₂were produced? The molar mass of CO₂ is 44.01 g/mol. Choose the closest answer.

1.991 g 0.4401 g 0.04523 g 3.981 g

Q2a. Based on the number of grams of CO₂ and H₂O produced during the reaction, does this compensate for the observed loss of mass?

A. No, because this violates the law of conservation of mass

B. yes, this compensates for the mass lost in the reaction

C. No, because more copper oxide was produced

D. No because mass is conserved

b. You observed that the products in the crucible weigh less than the reactants that you added. How is this possible?

A. Mass was lost due to experimental error

B. The Malachite absorbed CO2 from the air

C. CO2 and H2O were released as gases

D. Mass was converted to energy

c. What kind of a reaction took place upon heating?

A. decomposition

B. single displacement

C. synthesis

D. Double displacement

d. Baking soda is used as a leavening agent in baking. When heated, baking soda undergoes a decomposition reaction to form carbon dioxide. Given the data below, which of the following reactions occurs during baking? The molar mass of baking soda is 84.007 g/mol and the molar mass of carbon dioxide is 44.01 g/mol.

Mass of empty crucible (g) = 55.000

mass of crucible with baking soda (g) = 57.000

mass of crucible and contents after the reaction (g) =56.260

A. 2NaHCO3= 2Na+2CO2 +H2O

B. 2NaHCO3 = Na2O + 2CO2 +H2O

C. NaHCO3 = NaOH +CO2

D. 2NaHCO3 = Na2CO3 + CO2 +H2O

CHM 2046 Lecture Notes - Lecture 12: Strontium Hydroxide, Decimal Mark, Electronegativity

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