Lab Chemistry 107 (final ) Test

2.

Calculate the heat (KJ) evolved through the dissolved of 9.0 (g) NaOH in 90 (g) of water if the tempreture of the solution changed by 6.0 C⁰ (S.P=4.184 J/g.C⁰)

9485.3
2.5
3.5
3275.7

Correct Answer

A. 2.5

3.

A solution containing 0.840 Kg of CHCL₃ and 57.0 of eucaytol (MM=144) if T_f (CHCL₃) = -63.5 C⁰and K_f = 4.68 calculate the freezing point (C⁰ ) OF yhe solution

-67.7
-65.7
-66.7
-66.0
-61.3

Correct Answer

A. -65.7

Explanation

The freezing point of a solution is calculated using the formula: ΔTf = Kf * m.

Given that the freezing point depression constant (Kf) is 4.68 and the mass of the solution (m) is the sum of the mass of CHCL3 (0.840 kg) and eucalyptol (57.0 g), we can calculate the freezing point depression (ΔTf).

ΔTf = 4.68 * (0.840 + 0.057) = 4.68 * 0.897 = 4.1832

To find the freezing point (C0) of the solution, we subtract the freezing point depression from the freezing point of the pure solvent (Tf).

C0 = Tf - ΔTf = -63.5 - 4.1832 = -67.6832

Rounding to the nearest tenth, the freezing point of the solution is -65.7°C.

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4.

If 0.750 g of Cu 63.5 g/mol produced from the reaction of 15ml CuSO₄ with excess Zn , the molarity of CuSO₄ solution is :

0.99
0.88
0.79
0.72

Correct Answer

A. 0.79

Explanation

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this question, we are given the mass of Cu produced (0.750 g) and the molar mass of Cu (63.5 g/mol). By dividing the mass of Cu by its molar mass, we can find the number of moles of Cu produced. Since the reaction is stated to be with excess Zn, we can assume that all the CuSO4 reacted. The volume of the solution is given as 15 ml, which can be converted to liters by dividing by 1000. By dividing the number of moles of Cu by the volume of the solution, we can find the molarity of the CuSO4 solution, which is approximately 0.79 M.

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5.

Which of the following is most soluble in hot water

PbCl₂
AgCl
Hg₂Cl₂
PbCrO₄

Correct Answer

A. PbCl₂

Explanation

PbCl2 is the most soluble in hot water because it is a chloride compound, and chloride compounds tend to be more soluble in water compared to other compounds. Additionally, the presence of heat increases the kinetic energy of the particles, allowing them to overcome intermolecular forces and dissolve more readily.

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6.

The method used to separate PbCl₂from a mixture of AgCl and Hg₂Cl₂ after adding hot water to the mixture

Filtration
Sublimation
Distillation
Centrifugation

Correct Answer

A. Centrifugation

Explanation

Centrifugation is the correct answer because it is the most effective method to separate PbCl2 from a mixture of AgCl and Hg2Cl2 after adding hot water. Centrifugation involves spinning the mixture at high speeds, causing the heavier PbCl2 to settle at the bottom of the tube while the lighter AgCl and Hg2Cl2 remain suspended. This allows for easy separation of the PbCl2 by pouring off the supernatant liquid. Filtration would not be effective as the particles are too small to be trapped by the filter. Sublimation and distillation are not applicable in this scenario.

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7.

A 14 ml sample of H₂SO₄ solution required 25 ml of 0.2 M NaOH to neutralize...the molarity of H₂SO₄ is a

0.179
0.27857
0.17
0.2787

Correct Answer

A. 0.179

Explanation

The molarity of a solution can be calculated using the formula: Molarity = (moles of solute) / (volume of solution in liters). In this question, we know the volume of NaOH solution (25 mL) and its molarity (0.2 M). Since NaOH and H2SO4 react in a 1:1 ratio, the moles of H2SO4 can be calculated by multiplying the volume of NaOH solution (25 mL) by its molarity (0.2 M). The volume of the H2SO4 solution is given as 14 mL. To find the molarity of H2SO4, we divide the moles of H2SO4 by the volume of the H2SO4 solution in liters (14 mL = 0.014 L). Therefore, the molarity of H2SO4 is 0.2 M * 25 mL / 0.014 L = 0.357 M, which is equivalent to 0.179 when rounded to three decimal places.

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8.

Cu + 2Ag⁺ react to get Cu⁺² +2Ag

CU is oxidizing agent
Cu⁺² is reducing agent
Ag⁺is reducing agent
Cu is reducing agent

Correct Answer

A. Cu is reducing agent

Explanation

Cu is the reducing agent in this reaction because it undergoes oxidation, losing two electrons to form Cu+2. The Ag+ ions are the oxidizing agents because they gain electrons, being reduced to form Ag atoms.

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9.

The quantative yield of redox reaction which of the following was used to dissolve excess Zn (unreact):

NO₃
HCl
K₂CrO₄
H₂C₂H₂O₂

Correct Answer

A. HCl

Explanation

HCl is the correct answer because it is a strong acid that can dissolve excess Zn (unreacted) in a redox reaction. It can react with the Zn to form ZnCl2, which is soluble in water. This allows for the removal of the excess Zn from the reaction mixture, increasing the quantitative yield of the desired product.

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10.

What is the color which is produced from react Zn and CuSO₄ to get ZnSO₄ and Cu

Dark blue
Deep red
Dark blue
Brownish yellow

Correct Answer

A. Brownish yellow

Explanation

When zinc (Zn) reacts with copper sulfate (CuSO4), zinc sulfate (ZnSO4) and copper (Cu) are produced. The color of copper sulfate is dark blue, but when zinc is added to it, the color changes to brownish yellow. Therefore, the correct answer is brownish yellow.

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11.

One of the following is false :

Na oxidized by Fe⁺³
Ag⁺ reduced by Zn
Zn oxidized by Ag⁺
Au oxidized by Ag⁺

Correct Answer

A. Au oxidized by Ag⁺

Explanation

The given statement "Au oxidized by Ag+" is false. Gold (Au) is a noble metal and is not easily oxidized by other substances, including Ag+ (silver ion). In fact, gold is known for its resistance to oxidation and is often used in jewelry and other applications due to its inertness. Therefore, it is unlikely for gold to be oxidized by silver ions.

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12.

How many grams of Cu ( M.W=63.5 g/mol ) be produce through the reaction of 157 ml of .94 M CuSO₄ solution with excess of Zn

93.7133
9371.33
9.4
937.33

Correct Answer

A. 9.4

Explanation

The answer is 9.4 grams because the question states that there is an excess of Zn, meaning that Zn is not the limiting reactant. Therefore, the amount of Cu produced will be determined by the amount of CuSO4 present. To find the amount of CuSO4, we use the formula Molarity = moles/volume. Rearranging the formula to solve for moles, we get moles = Molarity * volume. Plugging in the values given, we get moles of CuSO4 = 0.94 mol/L * 0.157 L = 0.14758 mol. Since the molar ratio between CuSO4 and Cu is 1:1, the moles of Cu produced will also be 0.14758 mol. Finally, we can convert moles to grams using the molar mass of Cu (63.5 g/mol), giving us 9.36133 grams, which rounds to 9.4 grams.

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Lab Chemistry 107 (Final ) Test

The color of CuSo₄ is :

Quiz Preview

Calculate the heat (KJ) evolved through the dissolved of 9.0 (g) NaOH in 90 (g) of water if the tempreture of the solution changed by 6.0 C⁰ (S.P=4.184 J/g.C⁰)

A solution containing 0.840 Kg of CHCL₃ and 57.0 of eucaytol (MM=144) if T_f (CHCL₃) = -63.5 C⁰and K_f = 4.68 calculate the freezing point (C⁰ ) OF yhe solution

If 0.750 g of Cu 63.5 g/mol produced from the reaction of 15ml CuSO₄ with excess Zn , the molarity of CuSO₄ solution is :

Which of the following is most soluble in hot water

The method used to separate PbCl₂from a mixture of AgCl and Hg₂Cl₂ after adding hot water to the mixture

A 14 ml sample of H₂SO₄ solution required 25 ml of 0.2 M NaOH to neutralize...the molarity of H₂SO₄ is a

Cu + 2Ag⁺ react to get Cu⁺² +2Ag

The quantative yield of redox reaction which of the following was used to dissolve excess Zn (unreact):

What is the color which is produced from react Zn and CuSO₄ to get ZnSO₄ and Cu

One of the following is false :

How many grams of Cu ( M.W=63.5 g/mol ) be produce through the reaction of 157 ml of .94 M CuSO₄ solution with excess of Zn

Lab Chemistry 107 (Final ) Test

The color of CuSo4 is :

Quiz Preview

Calculate the heat (KJ) evolved through the dissolved of 9.0 (g) NaOH in 90 (g) of water if the tempreture of the solution changed by 6.0 C0 (S.P=4.184 J/g.C0)

A solution containing 0.840 Kg of CHCL3 and 57.0 of eucaytol (MM=144) if Tf (CHCL3) = -63.5 C0 and Kf = 4.68 calculate the freezing point (C0 ) OF yhe solution

If 0.750 g of Cu 63.5 g/mol produced from the reaction of 15ml CuSO4 with excess Zn , the molarity of CuSO4 solution is :

Which of the following is most soluble in hot water

The method used to separate PbCl2 from a mixture of AgCl and Hg2Cl2 after adding hot water to the mixture

A 14 ml sample of H2SO4 solution required 25 ml of 0.2 M NaOH to neutralize...the molarity of H2SO4 is a

Cu + 2Ag+ react to get Cu+2 +2Ag

The quantative yield of redox reaction which of the following was used to dissolve excess Zn (unreact):

What is the color which is produced from react Zn and CuSO4 to get ZnSO4 and Cu

One of the following is false :

How many grams of Cu ( M.W=63.5 g/mol ) be produce through the reaction of 157 ml of .94 M CuSO4 solution with excess of Zn

The color of CuSo₄ is :

Calculate the heat (KJ) evolved through the dissolved of 9.0 (g) NaOH in 90 (g) of water if the tempreture of the solution changed by 6.0 C⁰ (S.P=4.184 J/g.C⁰)

A solution containing 0.840 Kg of CHCL₃ and 57.0 of eucaytol (MM=144) if T_f (CHCL₃) = -63.5 C⁰and K_f = 4.68 calculate the freezing point (C⁰ ) OF yhe solution

If 0.750 g of Cu 63.5 g/mol produced from the reaction of 15ml CuSO₄ with excess Zn , the molarity of CuSO₄ solution is :

The method used to separate PbCl₂from a mixture of AgCl and Hg₂Cl₂ after adding hot water to the mixture

A 14 ml sample of H₂SO₄ solution required 25 ml of 0.2 M NaOH to neutralize...the molarity of H₂SO₄ is a

Cu + 2Ag⁺ react to get Cu⁺² +2Ag

What is the color which is produced from react Zn and CuSO₄ to get ZnSO₄ and Cu

How many grams of Cu ( M.W=63.5 g/mol ) be produce through the reaction of 157 ml of .94 M CuSO₄ solution with excess of Zn