MC: Ch. 4 pt. 2

How many grams of (NH4)3PO4 are needed to make 0.250 L of 0.150 M (NH4)3PO4?

(0.250 L) x (0.150 mol/L (NH4)3PO4) x (149.0867 g (NH4)3PO4/mol) = 5.59 g (NH4)3PO4 2.52 x 10-4 g *5.59 g 89.5 g 5.47 g

The value of ΔH° for the reaction below is -186 kJ. Calculate the heat (kJ) released from the reaction of 25 g of Cl2. H2 (g) + Cl2 (g) → 2HCl (g)

(25gCl₂) x (1 mol/70.954Cl₂)= 0.352 .352 x (-186kJ/1molCl₂)= heat released is -65.58 answer 66 kJ

The combustion of exactly 1.000 g of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat. If the combustion of 0.550 g of benzoic acid causes the temperature of the calorimeter to increase from 22.01∘C to 24.27∘C, calculate the heat capacity of the calorimeter.

(26.38 kJ) / (1.000 g) x (0.550 g) / (24.27∘C - 22.01∘C) = 6.42 kJ/∘C

The value of ΔH° for the reaction below is +128.1 kJ: CH3OH (l) → CO (g) + 2H2 (g) How many kJ of heat are consumed when 5.10 g of CO (g) is formed as shown in the equation?

(5.10gCO) x (1 mol/28.0101gCO)= 0.182 .182 x (128.1/1molCO)= heat consumed is 23.3 answer 23.3 kJ

The value of ΔH° for the reaction below is -1107 kJ: 2Ba (s) + O2 (g) → 2BaO (s) How many kJ of heat are released when 5.75 g of Ba (s) reacts completely with oxygen to form BaO (s)?

(5.75gBa) x (1 mol/137.327gBa)= .042 .042 x (-1107/2molBa)= heat released is -23.2 answer 23.3 kJ

The value of ΔH° for the reaction below is -1107 kJ: 2Ba (s) + O2 (g) → 2BaO (s) How many kJ of heat are released when 5.75 g of BaO (s) is produced?

(5.75gBa) x (1 mol/153.326gBaO)= 0.037 .037 x (-1107/2molBa)= heat released is -20.8 answer 20.8 kJ

The combustion of titanium with oxygen produces titanium dioxide: Ti (s) + O2(g) → TiO2 (s) When 2.060 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 91.60 °C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is ________ kJ/mol.

(91.60-25.0°C)=66.6°C 66.6°C x 9.84kJ/K = 655.344kJ 2.060g Ti/47.9g/mol Ti = 0.043 moles Ti per 0.043 mole, the kJ increase is 655.344kJ 655.344kJ/0.043 mole = 15,240.55 kJ/mol energy released (exothermic) so heat of combustion is negative -1.52x10⁴ kJ/mol

The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is ________.

*H+ (aq) + OH- (aq) → H2O (l) H+ (aq) + Na+ (aq) + OH- (aq) → H2O (l) + Na+ (aq) HNO3 (aq) + OH- (aq) → NO3- (aq) + H2O (l) H+ (aq) + HNO3 (aq) + 2OH- (aq) → 2H2O (l) + NO3- (aq) HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

Oxidation is the ________ and reduction is the ________.

*loss of electrons, gain of electrons gain of oxygen, loss of mass loss of oxygen, gain of electrons gain of electrons, loss of electrons gain of oxygen, loss of electrons

Calculate ΔH for 2 NO(g) + O2(g) → N2O4(g) using the following information: N2O4(g)→2 NO(g) ΔH==+57.9 2 NO2(g) = O2(g)→2 NO2(g) ΔH kJ−114.1 kJ

-57.9-114.1= -172kJ

What is the concentration of K+ in 0.15 M of K2S?

0.3 M

A stalk of celery has a caloric content (fuel value) of 9.0 kcal. If 1.0 kcal is provided by fat and there is very little protein, estimate the number of grams of carbohydrate and fat in the celery.

1 g carbohydrate and 2 g of fat 32 g carbohydrate and 10 g fat 2.2 g carbohydrate and 0.1 g fat 2 g carbohydrate and 1 g fat **2 g carbohydrate and 0.1 g fat

An orbital has n = 4 and ml = -1. What are the possible values of l for this orbital?

1,2,3 l is the angular momentum quantum number and it can have integral values from 0 to (n- 1) for each value of n. The ml, the magnetic quantum number, can have a value between -l to l. So, l cannot be 0 if ml is -1. The maximum value of l is 3 thus all possible values of l are 1, 2, 3.

What is the molarity of a NaOH solution if 15.5 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the NaOH solution?

1.15 0.573 0.220 *0.440 68.8

Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2. From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s) + 2H2O(l)→Ca(OH)2(s) + C2H2(g) ΔH∘=−127.2kJ

2H2O > 2H2 +O2 -H= 2* -285.83 Ca + O2 + H2 > Ca(OH)2 H= -986.2 2C + H2 > C2H2 H= 226.77 The above H's stand for standard enthalpy of formations. These can be found in textbook appendix. Notice the negative infront of the enthalpy (H) for H2O. This is to remind me/you that the heat lost is gained in the rxn. So then you add them up. 226.77 - 986.2 + (2*285.83) = -187.77 Add back the total enthalpy that is given in the question -187.77+127.2 = -60.57

Given the following reactions 2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -790 kJ S (s) + O2 (g) → SO2(g) ΔH = -297 kJ the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) + O2 (g) → 2SO3 (g) is ________ kJ.

2SO2 (g) + O2 (g) → 2SO3 (g) ORIGINAL EQUATION 2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -790 kJ because SO2 is on opposite side of arrow, you have to flip the equation and make ∆H positive also because there are 2 SO2 in OG multiple everything including ∆H by 2 (2)S (s) + (2)O2 (g) → (2)SO2(g) ΔH = (2)-297 kJ This equation↓ is the flipped/multipled by 2 version (2)SO2(g)→(2)S(s) + (2)O2(g) ∆H = +594 kJ -790 + 594 = -196 kJ 2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -790 kJ SO2(g)→S(s) + O2(g) ∆H = +297 kJ cross out same things in these two equations to make original equation answer is -196 kJ

The balanced molecular equation for complete neutralization of H2SO4 by KOH in aqueous solution is ________.

2SO4 (aq) + 2KOH (aq) → 2H2O (l) + K2SO4 (s) 2H+ (aq) + 2OH- (aq) → 2H2O (l) *H2SO4 (aq) + 2KOH (aq) → 2H2O (l) + K2SO4 (aq) 2H+ (aq) + 2KOH (aq) → 2H2O (l) + 2K+ (aq) H2SO4 (aq) + 2OH- (aq) → 2H2O (l) + SO42- (aq)

If you have an aqueous solution that contains 1.5 mol of HCl, how many moles of ions are in the solution?

3.0 mols HCl is a strong electrolyte that completely ionizes in water, thus 1.5 mol of hydrogen ions are formed and 1.5 mol of chloride ions are formed to give a total of 3.0 mol of ions when in aqueous solution.

How many milligrams of sodium sulfide are needed to completely react with 25.00 mL of a 0.0100 M aqueous solution of cadmium nitrate, to form a precipitate of CdS(s)?

3.8 mg 19.5 mg 23.5 mg 32.1 mg 39.0 mg The coefficients in a balanced equation determine the ratio of moles of reactants to moles of products. Na2S + Cd(NO3)2 → 2 NaNO3 + CdS According to the coefficients in the equation above, 1 mole of Na2S reacts with 1 mole of Cd(NO3)2 to produce 2 moles of NaNO3 + 1 mole of CdS. To determine the number of moles of Cd(NO3)2, use the following equations. Number of moles = Volume in liters * Molarity = 0.025 * 0.01 = 0.00025 Since the ratio is 1:1, the number of moles of Na2S is equal to the number of moles of Cd(NO3)2. To determine the mass of Na2s, multiply the number of moles by the mass of one mole. Mass of one mole = 2 * 23 + 32.1 = 78.1 Mass = 0.00025 * 78.1 = 0.019525 grams This is 19.525 mg.

What is the maximum number of electrons that can occupy each of the following subshells?

3p 6 electrons 5d 10 electrons 2s 2 electrons 4f 14 electrons The s orbitals Groups 1 & 2 (columns) can hold 2 electrons The p orbitals Groups 13 - 18 (columns) can hold 6 electrons The d orbitals Groups 3-12 (columns) can hold 10 electrons. The f orbitals can hold 14 electrons.

What are the respective concentrations (M) of Mg2+ and C2H3O2- afforded by dissolving 0.600 mol Mg(C2H3O2)2 in water and diluting to 135 mL?

4.44 and 8.89 0.0444 and 0.0889 0..889 and 0.444 0.444 and 0.889 0.444 and 0.444 Mg(C2H3O2)2 ---------> Mg2+ + 2C2H3O2- 1 mol mag acetate dissosiates into 1 mol Mg ion and 2 mols acetate ions in water therefore if 0.6 mol mag acetate is dissolved in 0.135 lits of water, we will have 0.6/0.135 = 4.44 moles / lit Mg2+ ions and 0.6*2/0.135 = 8.88 moles / lit C2H3O2- ions

At 20 ∘C (approximately room temperature) the average velocity of N2 molecules in air is 1050 mph. What is the average speed in m/s? What is the kinetic energy (in J) of an N2 molecule moving at this speed? What is the total kinetic energy of 1 mol of N2 molecules moving at this speed?

469.4m/s N₂= 14.0067 x 2= 28.0134g 28.0134g/6.02x10²³ = 4.65x10⁻²³ Ek= 1/2 (4.65x10⁻23)(469m/s)² Ek= 5.124 x 10⁻²¹J 5.124x1-⁻²¹J/molecule x 6.02x10²³ molecules/mol = 3.084.648J = 3.086kJ/mol

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0% H2SO4 by mass and has a density of 1.84 g/mL. Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution?

6210 kg Balanced equation: H2SO4 + Na2CO3 → Na2SO4 + CO2 + H2O 1mol H2SO4 reacts with 1mol Na2CO3 Molar mass H2SO4 = 98.08 g/mol Therefore you have 5747.5*1000 / 98.08 = 58,600.12moles of H2SO4 You need the same moles of Na2CO3 = 58,600.12mol Molar mass Na2CO3 = 105.9884 g/mol (anhydrous)124.00 g/mol (monohydrate) and 286.14 g/mol (decahydrate) For simplicity we will work with the anhydrous product: 58,600.12 * 105.9884/1000 = 6,210.9kg Na2CO3 anhydrous required.

Which of the following objects has the greatest kinetic energy?

A) a 500-kg motorcycle moving at 100 km/h B) a 1,000-kg car moving at 50 km/h C) a 1,500-kg car moving at 30 km/h D) a 5,000-kg truck moving at 10 km/h E) a 10,000-kg truck moving at 5 km/h A because Ek= 1/2mv²

The net ionic equation for formation of an aqueous solution of Al(NO3)3 via mixing solid Al(OH)3 and aqueous nitric acid is ________.

Al(OH)3 (s) + 3HNO3 (aq) → 3H2O (l) + Al(NO3)3 (aq) Al(OH)3 (s) + 3NO3- (aq) → 3OH- (aq) + Al(NO3)3 (aq) Al(OH)3 (s) + 3NO3- (aq) → 3OH- (aq) + Al(NO3)3 (s) *Al(OH)3 (s) + 3H+ (aq) → 3H2O (l) + Al3+ (aq) Al(OH)3 (s) + 3HNO3 (aq) → 3H2O (l) + Al3+ (aq) + NO3- (aq)

Which of the following is an oxidation-reduction reaction?

Ba(C2H3O2)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaC2H3O2 (aq) HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq) H2CO3 (aq) + Ca(NO3)2 (aq) → 2HNO3 (aq) + CaCO3 (s) AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq) *Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq)

Which element is oxidized in this reaction? 2CuO+C→2Cu+CO₂ Which substance is the oxidizing agent in this reaction? 2CuO+C→2Cu+CO₂ Which element is reduced in this reaction? 2KMnO₄+3Na₂SO₃+H2O→2MnO₂+3Na₂SO₄+2KOH Which substance is the reducing agent in this reaction? 2KMnO₄+3Na₂SO₃+H₂O→2MnO₂+3Na₂SO₄+2KOH

C. Carbon gets oxidized in the process of reducing copper. Copper gets reduced in the process of oxidizing carbon. CuO. Notice that the oxidizing agent is itself reduced. The species oxidized is not the oxidizing agent, but the reducing agent. Think of the oxidizing agent as the species the acts upon another species causing oxidation. Mn. Sulfur gets oxidized in the process of reducing manganese. Manganese gets reduced in the process of oxidizing sulfur. Na₂SO₃. The reducing agent acts to reduce another species and therefore must itself be oxidized.

Alexander, who weighs 186 lb , decides to climb Mt. Krumpett, which is 5520 m high. For his food supply, he decides to take nutrition bars. The label on the bars states that each 100-g bar contains 10 g of fat, 40 g of protein, and 50 g of carbohydrates. One gram of fat contains 9 Calories, whereas each gram of protein and carbohydrates contains 4 Calories. Nutrition facts Calories per gram Fat 9 Protein 4 Carbohydrates 4 Alexander wants to know exactly how many bars to pack in his backpack for the journey. To provide a margin of safety, he assumes that he will need as much energy for the return trip as for the uphill climb. How many bars should Alexander pack?

E= mgh (84.3682kg)(9.81m/s)²(5520m)= 4568639.27184 Joules x 2= 9137278.54368 Joules Joules to calories 2183.86 kcal 10g * 9kcal/g +40g * 4kcal/g + 50g * 4kcal/g = 450 kcal per bar. 2183.86 kcal/450 kcal= 4.85 bars

If the heat of formation of H2O(l) is -286 kJ/mol, which of the following thermochemical equations for the formation of H2O is correct?

H2(g) + O(g) → H2O(g) ΔH=−286 kJ **H2(g) + 1/2 O2(g) → H2O(l) ΔH=−286 kJ 2 H2(g) + O2(g) → 2 H2O(l) ΔH=−286kJ 2 H(g) + O(g) → H2O(l) ΔH=−286 kJ H2O(l) → H2(g) + 1/2 O2(g) ΔH=286 kJ

An aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of NaCl of the same concentration. Which of the following substances could the unknown be: KOH, NH3, HNO3, KClO2, H3PO3, CH3COCH3 (acetone)?

H3PO3 is a weak acid and also forms ions in aqueous solutions.

Formic acid, HCOOH, is a weak electrolyte. What solute particles are present in an aqueous solution of this compound? Write the chemical equation for the ionization of HCOOH.

HCOOH,H+,HCOO− HCOOH(aq)⇌H+(aq)+HCOO−(aq)

In which species does chlorine have the highest oxidation number?

HClO₄ the oxidation state of Cl gets higher as more O is added

Consider the following three moving objects: A golf ball with a mass of 45.9 g moving at a speed of 50.0 m/s. An electron moving at a speed of 3.5×105 m/s. A neutron moving at a speed of 2.3×102 m/s. List the three objects in order from shortest to longest de Broglie wavelength.

If you calculate the de Broglie wavelength for each object, you should obtain the following: The de Broglie wavelength of the golf ball is 2.89 × 10−34m. The de Broglie wavelength of the electron is 21 Å. The de Broglie wavelength of the neutron 17 Å. The de Broglie wavelength is inversely proportional to the momentum. So, the order from shortest to longest de Broglie wavelength is i < iii < ii.

If CaCl2 is dissolved in water, what can be said about the concentration of the Ca2+ ion?

It has the same concentration as the Cl− ion. *Its concentration is half that of the Cl− ion. Its concentration is twice that of the Cl− ion. Its concentration is one-third that of the Cl− ion.

What dissolved species are present in a solution of KCN? What dissolved species are present in a solution of NaClO4?

K+,CN− Na+,ClO4−

kinetic energy

Kinetic energy, Ek, is given by the formula Ek=12mv2 where m is mass is the mass in kilograms and v is velocity in meters per second.

A certain atom has an ns2np6 electron configuration in its outermost occupied shell. Which of the following elements could it be?

Kr ns2np6 electron configuration indicates eight valence electrons. Krypton, Kr has eight electrons its outermost occupied shell. The electron configuration for Kr is [Ar] 3d104s24p6

One cup of fresh orange juice contains 136 mg of ascorbic acid (vitamin C, C6H8O6). Given that one cup = 260.0 mL calculate the molarity of vitamin C in organic juice.

M = 2.97×10−3 M mol = mass/molar mass = 0.136/176.12 = 7.72 * 10^-4 Molarity = mol/Volume(L) = 7.72 * 10^-4/0.260 =2.97x10-³M(mol/L)

What is the net ionic equation of the reaction of MgCl2 with NaOH?

Mg2+(aq)+2OH−(aq)→Mg(OH)2(s) the balanced equation is MgCl2 + 2NaOH -----> Mg(OH)2 + 2NaCl Write the complete ionic equation from the balanced equation by noting that the soluble compounds MgCl2, NaOH, and NaCl exist as ions. Mg2+(aq) + 2Cl-(aq) + Na+(aq) + OH-(aq) -----> Mg(OH)2(s) + Na+(aq) + Cl-(aq) Then, find the net ionic equation by cancelling out any species that appear on both sides of the equation. Mg2+(aq) + 2OH-(aq) -----> Mg(OH)2(s)

Given the following reactions N2 (g) + 2O2 (g) → 2NO2 (g) ΔH = 66.4 kJ 2NO (g) + O2 (g) → 2NO2 (g) ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen to produce nitric oxide N2 (g) + O2 (g) → 2NO (g) is ________ kJ.

N2 (g) + O2 (g) → 2NO (g) ORIGINAL N2 (g) + 2O2 (g) → 2NO2 (g) ΔH = 66.4 kJ flip this equation because 2NO is on opposite sides, so ∆H sign changes. 2NO (g) + O2 (g) → 2NO2 (g) ΔH = -114.2 kJ Now becomes ∆H= 114.2 66.4kJ + 114.2kJ= 180.6kJ

Given the following reactions 2NO → N2 + O2 ΔH = -180 kJ 2NO + O2 → 2NO2 ΔH = -112 kJ the enthalpy of the reaction of nitrogen with oxygen to produce nitrogen dioxide N2 + 2O2 → 2NO2 is ________ kJ.

N2 + 2O2 → 2NO2 ORIGINAL 2NO → N2 + O2 ΔH = -180 kJ 2NO + O2 → 2NO2 ΔH = -112 kJ flip both equations because 2NO is on opposite sides for each compared the OG, so ∆H sign will change. 180kJ 112kJ 180kJ-112kJ= 68kJ

Given the following data: N2(g) + O2(g) →2NO(g) ΔH= +180.7 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH= -113.1 kJ 2N2O(g) → 2N2(g) + O2(g) ΔH= -163.2 kJ use Hess's law to calculate ΔH for the following reaction: N2O(g)+NO2(g)→3NO(g)

N2O(g)+NO2(g)→(2)3NO(g) ORIGINAL N2(g) + O2(g) →2NO(g) ΔH= (2)+180.7 kJ (same side, x2) 2NO(g) + O2(g) → 2NO2(g) ΔH= -113.1 kJ (opposite side, inverted, sign change) 2N2O(g) → 2N2(g) + O2(g) ΔH= -163.2 kJ (no change) 2N2 + 2O2→4NO ∆H= 361.4 kJ 2NO2 → 2NO + O2 ∆H= 113.1 kJ 2N2O → 2N2 + O2 ∆H= -163.2 kJ now add them 2N2 + 2O2 + 2NO2 + 2N2O→ 6NO + 2N2 + 2O2 361.4 kJ + 113.1 kJ + -163.2 kJ= 311.3 kj cancel terms 2NO2 + 2N2O→6NO ∆H= 311.3 kJ factorize by 2 NO2 + N2O → 3NO ∆H= 155.65

Will a precipitate form when solutions of Ba(NO3)2 and KOH are mixed?

NO Ba(NO3)2 and KOH mix to form Ba(OH)2 and KNO3 which are both soluble in water thus no precipitate will be formed.

What are the spectator ions in the reaction between KOH (aq) and HNO3 (aq)?

OH- only H+ and NO3- *K+ and NO3- H+ and OH- K+ and H+

The temperature of a 15-g sample of lead metal increases from 22 °C to 37 °C upon the addition of 29.0 J of heat. The specific heat capacity of the lead is ________ J/g-K.

Q= mc∆T 29.0J= (15g)(X)(37-22) 29.0J = 225X X= 29.0J/225= 0.128J/g-K.

What is the net ionic equation of the reaction of MgSO4 with Pb(NO3)2?

SO₄²−(aq)+Pb²+(aq)→PbSO₄(s) The balanced equation for this reaction is MgSO₄(aq)+Pb(NO₃)₂(aq)→Mg(NO₃)₂(aq)+PbSO₄(s) The complete ionic equation is Mg²+(aq)+SO₄²−(aq)+Pb²+(aq)+2NO³-(aq)→PbSO₄(s)+Mg²+(aq)+2NO³-(aq) If you eliminate spectator ions, which are the ions thatcan be found both on the reactants' and on the products' side, you'll get the net ionic equation SO2−4(aq)+Pb2+(aq)→PbSO4(s)

Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Hydrochloric acid HCl Sodium hydroxide NaOH Carbonic acid H2CO3 Ethyl amine CH3CH2NH2 Potassium iodide KI Propanol C3H7OH Sucrose C12H22O11

Strong electrolyte Hydrochloric acid HCl Potassium iodide KI Sodium hydroxide NaOH weak electrolyte Carbonic acid H2CO3 Ethyl amine CH3CH2NH2 nonelectrolyte Propanol C3H7OH Sucrose C12H22O11

When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, the following reaction occurs: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) If the temperature of the solution increases from 23.0 ∘C to 34.1 ∘C as a result of this reaction, calculate ΔH in kJ/mol of Mg. Assume that the solution has a specific heat of 4.18 J/g∘C.

Supposing the solution's density to be the same as that of water: (4.18 J/g∘C) x (100 g) x (34.1 - 23.0)∘C = 4639.8 J (0.243 g Mg) / (24.30506 g Mg/mol) = 0.0099979 mol Mg (4.6398 kJ) / (0.0099979 mol) = 464 kJ/mol Since the reaction is exothermic, the ΔH is negative by convention, so: ΔH = - 464 kJ/mol

Relating Heat and Work to Changes of Internal Energy Consider the following four cases A) A chemical process in which heat is absorbed. B) A change in which q=30 J, w=44 J. C) A process in which a system does work on its surroundings with no change in q. D) A process in which work is done on a system and an equal amount of heat is withdrawn.

The internal energy changes for each case can be summarized as follows: A) A chemical process in which heat is absorbed results in an increase in internal energy. B) A change in which q=30 J, w=44 J. Since the change in internal energy is the sum of the heat added to or liberated from the system, q, and the work done on or by the system, w, this change results in an increase in internal energy. C) A process in which a system does work on its surroundings with no change in q results in a decrease in the internal energy because the systems is essentially losing energy to the surroundings. D) A process in which work is done on a system and an equal amount of heat is withdrawn has no net change in the internal energy. Thus out of the four processes, only (iii) results in a decrease in internal energy.

A mysterious white powder is found at a crime scene. A simple chemical analysis concludes that the powder is a mixture of sugar and morphine (C17H19NO3), a weak base similar to ammonia. The crime lab takes 10.00 mg of the mysterious white powder, dissolves it in 100.00 mL water, and titrates it to the equivalence point with 2.84 mL of a standard 0.0100 M HCl solution. What is the percentage of morphine in the white powder?

The number of moles of morphine that reacted with HCl was calculated by assuming a 1:1 stoichiometric ratio. Thus, the moles of morphine at the equivalence point was equal to the moles of HCl (0.0100 M×0.00284 L = 2.84×10−5 mol). The mass of morphine reacted was found by multiplying the moles reacted by the molar mass of morphine (0.008094 g). To determine the percentage of morphine in the white powder, the mass of morphine reacted was divided by the total mass of the white sample 0.008094g0.010g ×100 = 81.0%

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)? The combustion of ethene, C2H4, occurs via the reaction C2H4(g)+3O2(g)→2CO2(g)+2H2O(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) C2H4 (g) 52.47 CO2(g) −393.5 H2O(g) −241.8 Calculate the enthalpy for the combustion of 1 mole of ethene.

The standard heat of formation, ΔH∘f, is defined as the enthalpy change for the formation of ONE MOLE of substance from its constituent ELEMENTS in their standard states (not compounds). **Na(s)+12Cl2(g)→NaCl(s) CO(g)+12O2(g)→CO2(g) 2Na(s)+Cl2(g)→2NaCl(s) **C(s,graphite)+O2(g)→CO2(g) Na(s)+12Cl2(l)→NaCl(s) BaCO3(s)→BaO(s)+CO2(g) Thus, elements in their standard states have ΔH∘f=0. Heat of formation values can be used to calculate the enthalpy change of any reaction. ΔH∘rxn===ΔH∘f(products) C2H4 + 3 O2 → 2 CO2 + 2 H2O(g) "products - reactants", all in kJ/mol: (2 x −393.5) + (2 x −241.8) − (52.47) − (3 x 0) = −1323 kJ C2H4

How many of the elements in the second row of the periodic table (Li through Ne) will have at least one unpaired electron in their electron configurations?

There are six elements in the second row of the periodic table that have at least one unpaired electron in their electron configurations: Li, B, C, N, O, F.

What happens when you mix an aqueous solution of sodium nitrate with an aqueous solution of barium chloride?

There is no reaction; all possible products are soluble.

The complete combustion of ethanol, C2H5OH (FW = 46.0 g/mol), proceeds as follows: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = −555 kJ What is the enthalpy change for combustion of 15.0 g of ethanol?

To determine the enthalpy change for combustion of 15.0 g of ethanol follow these steps: determine moles of ethanol in 15 g : (mass ethanol/FW ethanol) 15.0 g/46.0 g = 0.3261 mol) find the enthalpy change for 0.3261 mol of ethanol: ΔH = -555 kJ for the combustion of 1 mol of ethanol, therefore for 0.3261 mol of ethanol −555 kJ/mol ethanol×0.03261 mol ethanol=−181 kJ.

How much work is done when a 195 g tomato is lifted 16.0 m ? The tomato is dropped. What is the velocity, v, of the tomato when it hits the ground? Assume 81.8 % of the work done in Part A is transferred to kinetic energy, E, by the time the tomato hits the ground.

W=Fd F=ma F= .195kg x 9.8m/s²= 1.911N W= 1.911N x 16.0m= 30.576N or 30.6J 1J=1N The energy transferred to the tomato by doing work on it increases its potential energy because the tomato is now at a greater height or position relative to the ground. Ep= mgh (.195kg)(9.8m/s)(16.0m)= 30.576J Ek= Ep x % 30.576J x .818= 25.0111J v=√2Ek/m = 16m/sec Once the tomato hits the ground, it stops moving and its kinetic energy goes to zero. Some of the kinetic energy is transferred into the work involved in smashing the tomato, while the rest is dissipated to the surroundings as heat.

Enthalpy diagram for combustion of 1 mol of methane. The enthalpy change of the one-step reaction equals the sum of the enthalpy changes of the reaction run in two steps: -890 kJ = -802 kJ+ (-88 kJ). image

What process corresponds to the -88 kJ enthalpy change? **A)The condensation of 2 H2O(g) to 2 H2O(l). B) The evaporation of 2 H2O(l) to 2 H2O(g). C) The reaction between CO2(g) and 2H2O(g).

Consider the following data for five hypothetical elements: Q, W, X, Y, and Z. Rank the elements from most reactive to least reactive. Combination Observation of reaction Q+W+ Reaction occurs X+Z+ No reaction W+Z+ Reaction occurs Q++Y Reaction occurs

Y,Q,W,Z,X

Oxidation and ________ mean essentially the same thing.

corrosion

Strong electrolytes _____.:

do not conduct electricity in solution. do not form ions in solution. *exist in solution completely or nearly completely as ions. exist in solution mostly in the form of neutral molecules with only a small fraction in the form of ions.

The point in a titration at which the indicator changes is called the ________.

endpoint

Which one of the following is a diprotic acid?

hydrofluoric acid *sulfuric acid chloric acid phosphoric acid nitric acid

Which of these metals is the easiest to oxidize?

lithium Although aluminum, sodium and iron can all be oxidized, lithium is the easiest to oxidized because it is the most active metal. Recall that the more active the metal the easier it will be to oxidize it. Table 4.5 in the book gives the activity series for metals in aqueous solution.

What is the name of the base that has the formula Mg(OH)2? What is the chemical formula for the base sodium hydroxide? Which of these compounds is not a base?

magnesium hydroxide NaOH HBr

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 475 mL of a solution that has a concentration of Na+ ions of 1.00 M ?

mass of Na3PO4 = 26.0 g Na3PO4 dissociates in 3 Na+ and PO43- concentration Na3PO4 = 1.00 M/ 3 = 0.333 M Moles Na3PO4 = 0.333 x 0.475 L=0.158 Mass Na3PO4 = moles x molar mass = 0.241 mol x 163.94 g/mol= 25.9g =26.0g

What is the concentration of ammonia in a solution made by dissolving 3.75 g of ammonia in 120.0 L of water?

mol = mass/m.wt = 3.75/17.03 = 0.22 mol Molarity = mol / Volume(L) = 0.22/120.0 = 1.84 * 10^-3 M 3.78×10−2 M 7.05 M *1.84×10−3 M 0.0313 M 1.84 M

When a 6.50-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (the following Figure), the temperature rises from 21.6 ∘C to 37.8 ∘C. Calculate ΔH (in kJ/mol NaOH) for the solution process NaOH(s)→Na+(aq)+OH−(aq) Assume that the specific heat of the solution is the same as that of pure water.

q= (100.0 + 6.50)g x 4.18J/g/C x (37.8-21.6C)= q= 7211.754J = 7.21kJ q released = −q absorbed q released = −7.21 kJ 6.50g x (1mol NaOH/40.0g)= 0.1625 moles NaOH ∆H = -7.21kJ/0.1625 moles NaOH = -44.4 kJ/mol

Consider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beakers in which a chemical reaction will occur and those in which no reaction will occur. Mn(s)+Ca(NO3)2(aq) KOH(aq)+Fe(s) Pt(NO3)2(aq)+Cu(s) Cr(s)+H2SO4(aq)

reaction no reaction C,D A,B

Oxidation cannot occur without ________.

reduction

From the enthalpies of reaction 2C(s)+O2(g)→2CO(g)ΔH=−221.0kJ 2C(s)+O2(g)+4H2(g)→2CH3OH(g)ΔH=−402.4kJ calculate ΔH for the reaction CO(g)+2H2(g)→CH3OH(g)

so you have 2CO but you only need one, plus its on the products side but it should be in reactant, so -221.0 becomes +221.0 and you divide by 2 = 110.5 You have 2 CH3OH and 4H2 but both should be half of that so -402.4/2 = -201.2 Now 110.5+(-201.2) = =90.7 answer = -90.7 kJ

Calculating Frequency from Wavelength Consider the following three statements: -For any electromagnetic radiation, the product of the wavelength and the frequency is a constant. -If a source of light has a wavelength of 3.0 Å, its frequency is 1.0×1018 Hz. -The speed of ultraviolet light is greater than the speed of microwave radiation. Which of these three statements is or are true?

statements 1 and 2 are true The only false statment is (3), "The speed of ultraviolet light is greater than the speed of microwave radiation" is false because the speed of any electromagnetic radiation is constant regardless the type of radiation.

Each of the following reactions shows a solute dissolved in water. Classify each solute as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. C(l)→C(aq) AB(aq)⇌A+(aq)+B−(aq) MN(aq)→M+(aq)+N−(aq) XZ(s)→X+(aq)+Z−(aq) P(s)→P(aq)

strong electrolyte XZ and MN weak electrolyte AB nonelectrolyte C and P

A piston has an external pressure of 10.0 atm. How much work has been done in joules if the cylinder goes from a volume of 0.130 liters to 0.450 liters?

w=-P(∆v) 1L x atm= 101.3J w= -10 x (.32) = -3.2 -3.2 x 101.3J = -324J

If a balloon is expanded from 0.055 L to 1.403 L against an external pressure of 1.02 atm, how many L⋅atm of work is done?

w=-P(∆v) w= -(1.02atm)(1.403L-0.055L) = -1.37Latm

A neutralization reaction between an acid and a metal hydroxide produces ________.

water and a salt

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 200 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

ΔH = -1.55 kJ/mol (4.184 J/g·°C) x (200 g) x (24.70 - 21.00)∘C = 3096.16 J = 3.09 kJ (3.09 kJ) / (2.00 mol) = -1.55 kJ/mol

A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When 0.440 kJ of heat is added to the gas, it expands and does 218 J of work on the surroundings. What is the value of ΔH for this process? What is the value of ΔE for this process?

∆H=0.440kJ ∆E= 440J-218J= 222J or .222kJ