Quant Chemistry (MCQ Test)
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9.15 g āĻŦāĻŋāĻļā§āĻĻā§āĻ§ ZnO āĻšāĻ¤ā§ āĻāĻžāĻ°ā§āĻŦāĻ¨ āĻŦāĻŋāĻāĻžāĻ°āĻŖ āĻĒāĻĻā§āĻ§āĻ¤āĻŋāĻ¤ā§ āĻāĻ¤ āĻā§āĻ°āĻžāĻŽ Zn āĻĒāĻžāĻā§āĻž āĻ¯āĻžā§?
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3.75
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9.75
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7.95
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7.35
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About This Quiz
This 'Quant Chemistry' quiz assesses knowledge in quantitative chemical analysis, focusing on reactions, stoichiometry, and electron transfer in various compounds. It is designed for learners to apply theoretical concepts practically, enhancing their understanding of chemical properties and reactions.
Quiz Preview
- 2.
āĻāĻ¸āĻŋāĻĄ āĻŽāĻžāĻ§ā§āĻ¯āĻŽā§ āĻĒāĻāĻžāĻļāĻŋā§āĻžāĻŽ āĻĒāĻžāĻ° āĻŽā§āĻ¯āĻžāĻā§āĻāĻžāĻ¨ā§āĻ āĻāĻ° āĻā§āĻšā§āĻ¤ āĻāĻ˛ā§āĻāĻā§āĻ°āĻ¨ āĻ¸āĻāĻā§āĻ¯āĻž-
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2
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3
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5
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7
Correct Answer
A. 5 -
- 3.
5g FeSO4 āĻā§ āĻ āĻŽā§āĻ˛ā§ā§ āĻŽāĻžāĻ§ā§āĻ¯āĻŽā§ āĻāĻžāĻ°āĻŋāĻ¤ āĻāĻ°āĻ¤ā§ āĻāĻ¤ āĻā§āĻ°āĻžāĻŽ KMnO4 āĻĒā§āĻ°ā§ā§āĻāĻ¨?
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1.04
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2.04
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4.01
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4.02
Correct Answer
A. 1.04 -
- 4.
āĻŽā§āĻĻā§ āĻāĻ¸āĻŋāĻĄ āĻ āĻ¤ā§āĻŦā§āĻ° āĻā§āĻˇāĻžāĻ°ā§āĻ° āĻŦāĻŋāĻā§āĻ°āĻŋā§āĻžā§ āĻāĻĒāĻ¯ā§āĻā§āĻ¤ āĻ¨āĻŋāĻ°ā§āĻĻā§āĻļāĻ āĻā§āĻ¨āĻāĻŋ?
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āĻĢā§āĻ¨āĻĢāĻĨā§āĻ¯āĻžāĻ˛āĻŋāĻ¨
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āĻŽāĻŋāĻĨāĻžāĻāĻ˛ āĻ āĻ°ā§āĻā§āĻ
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āĻŽāĻŋāĻĨāĻžāĻāĻ˛ āĻ°ā§āĻĄ
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āĻ¯ā§ āĻā§āĻ¨ āĻ¨āĻŋāĻ°ā§āĻĻā§āĻļāĻ
Correct Answer
A. āĻĢā§āĻ¨āĻĢāĻĨā§āĻ¯āĻžāĻ˛āĻŋāĻ¨Explanation
āĻĢā§āĻ¨āĻĢāĻĨā§āĻ¯āĻžāĻ˛āĻŋāĻ¨ āĻšāĻ˛ āĻāĻāĻāĻŋ āĻŽā§āĻĻā§ āĻāĻ¸āĻŋāĻĄ āĻ āĻ¤ā§āĻŦā§āĻ° āĻā§āĻˇāĻžāĻ°ā§āĻ° āĻŦāĻŋāĻā§āĻ°āĻŋā§āĻžā§ āĻāĻĒāĻ¯ā§āĻā§āĻ¤ āĻ¨āĻŋāĻ°ā§āĻĻā§āĻļāĻāĨ¤ āĻāĻāĻŋ āĻāĻ¸āĻŋāĻĄ-āĻŦā§āĻ¸ āĻ¨āĻŋāĻ°ā§āĻĻā§āĻļāĻ āĻšāĻŋāĻ¸ā§āĻŦā§ āĻŦā§āĻ¯āĻŦāĻšāĻžāĻ° āĻāĻ°āĻž āĻšā§āĨ¤ āĻāĻ° āĻŽāĻžāĻ§ā§āĻ¯āĻŽā§ āĻāĻ¸āĻŋāĻĄ āĻ āĻā§āĻˇāĻžāĻ°ā§āĻ° āĻĒāĻ°āĻŋāĻŽāĻžāĻŖ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĻž āĻ¯āĻžā§āĨ¤ āĻāĻāĻŋ āĻ°āĻā§āĻ° āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤āĻ¨ āĻāĻ°ā§ āĻāĻ¸āĻŋāĻĄ āĻ āĻā§āĻˇāĻžāĻ°ā§āĻ° āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĻžāĻ° āĻāĻ¨ā§āĻ¯ āĻŦā§āĻ¯āĻŦāĻšā§āĻ¤ āĻšā§āĨ¤ āĻ¸āĻŽā§āĻĒā§āĻ°ā§āĻŖ āĻŦāĻžāĻā§āĻ¯āĻāĻŋāĻ° āĻ āĻ°ā§āĻĨ āĻšāĻ˛, āĻĢā§āĻ¨āĻĢāĻĨā§āĻ¯āĻžāĻ˛āĻŋāĻ¨ āĻšāĻ˛ āĻŽā§āĻĻā§ āĻāĻ¸āĻŋāĻĄ āĻ āĻ¤ā§āĻŦā§āĻ° āĻā§āĻˇāĻžāĻ°ā§āĻ° āĻŦāĻŋāĻā§āĻ°āĻŋā§āĻžā§ āĻāĻĒāĻ¯ā§āĻā§āĻ¤ āĻ¨āĻŋāĻ°ā§āĻĻā§āĻļāĻāĨ¤Rate this question:
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- 5.
Ca(OCl)Cl āĻ āĻŖā§āĻ¤ā§ Cl āĻāĻ° āĻāĻžāĻ°āĻŖ āĻ āĻŦāĻ¸ā§āĻĨāĻž-
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+1
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-1
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0
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+1, -1
Correct Answer
A. +1, -1Explanation
The oxidation state of Cl in Ca(OCl)Cl is +1 and -1. In the compound, the oxidation state of Ca is +2, and since the overall charge of the compound is 0, the sum of the oxidation states of all the elements must be 0. Therefore, the oxidation state of Cl must be such that when added to the oxidation state of Ca (+2), it gives a sum of 0. Since there are two Cl atoms, one Cl atom must have an oxidation state of +1 and the other Cl atom must have an oxidation state of -1 in order to satisfy this condition.Rate this question:
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- 6.
0.1 M āĻ¸āĻŽāĻā§āĻ¤āĻ¨ā§āĻ° NaOH āĻ H2SO4 āĻāĻ° āĻĻā§āĻ°āĻŦāĻŖā§āĻ° āĻĒā§āĻ°āĻā§āĻ¤āĻŋ āĻāĻŋāĻ°ā§āĻĒ āĻšāĻŦā§?
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āĻāĻāĻ§āĻ°ā§āĻŽā§
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āĻ¨āĻŋāĻ°āĻĒā§āĻā§āĻˇ
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āĻ āĻŽā§āĻ˛ā§ā§
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āĻā§āĻˇāĻžāĻ°ā§ā§
Correct Answer
A. āĻ āĻŽā§āĻ˛ā§ā§Explanation
The nature of the solution formed by mixing 0.1 M NaOH and H2SO4 will be acidic because H2SO4 is a strong acid and NaOH is a strong base. When they react, they will form H2O and Na2SO4, which is a salt. The presence of H2SO4 in the solution will make it acidic.Rate this question:
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- 7.
1.032 g O2 āĻ 0.573g CO2 āĻŽāĻŋāĻļā§āĻ°āĻŖā§āĻ° CO2 āĻāĻ° āĻŽā§āĻ˛ āĻāĻā§āĻ¨āĻžāĻāĻļ āĻāĻ¤?
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0.713
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0.8323
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0.287
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0.1677
Correct Answer
A. 0.287Explanation
The molar mass of O2 is 32 g/mol and the molar mass of CO2 is 44 g/mol. To find the mole fraction of CO2 in the mixture, we need to calculate the moles of CO2 and the total moles of the mixture.
The moles of O2 can be calculated by dividing its mass by its molar mass:
moles of O2 = 1.032 g / 32 g/mol = 0.03225 mol
The moles of CO2 can be calculated in the same way:
moles of CO2 = 0.573 g / 44 g/mol = 0.01302 mol
The total moles of the mixture is the sum of the moles of O2 and CO2:
total moles = 0.03225 mol + 0.01302 mol = 0.04527 mol
Finally, the mole fraction of CO2 can be calculated by dividing the moles of CO2 by the total moles:
mole fraction of CO2 = 0.01302 mol / 0.04527 mol = 0.287Rate this question:
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- 8.
āĻ¨āĻŋāĻā§āĻ° āĻā§āĻ¨ āĻā§āĻ¨ āĻāĻžāĻ°āĻ āĻ āĻŦāĻŋāĻāĻžāĻ°āĻ āĻāĻā§ āĻšāĻŋāĻ¸ā§āĻŦā§ āĻāĻžāĻ āĻāĻ°ā§?
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Na+
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Fe2+
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Al3+
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Sn4+
Correct Answer
A. Fe2+Explanation
Fe2+ can function as both a reducing agent and an oxidizing agent. As a reducing agent, it can donate electrons to other substances and get oxidized in the process. As an oxidizing agent, it can accept electrons from other substances and get reduced in the process. This ability to undergo both reduction and oxidation reactions makes Fe2+ versatile and capable of acting as both a reducing agent and an oxidizing agent.Rate this question:
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- 9.
āĻŦāĻŋā§āĻžāĻ° āĻ˛ā§āĻ¯āĻžāĻŽā§āĻŦāĻžāĻ°āĻ āĻ¸ā§āĻ¤ā§āĻ° āĻā§āĻ¨ āĻŽā§āĻ˛āĻžāĻ° āĻĻā§āĻ°āĻŦāĻŖā§āĻ° āĻā§āĻˇā§āĻ¤ā§āĻ°ā§ āĻ āĻ§āĻŋāĻ āĻĒā§āĻ°āĻ¯ā§āĻā§āĻ¯?
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0.01
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0.1
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0.5
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1
Correct Answer
A. 0.01Explanation
The correct answer is 0.01 because the Beer-Lambert law states that the absorbance of a substance is directly proportional to the concentration of the substance in the solution. Therefore, a smaller value of absorbance (0.01) indicates a lower concentration of the solution, making it more applicable in the context of the Beer-Lambert law.Rate this question:
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- 10.
āĻŽāĻŋāĻĨāĻžāĻāĻ˛ āĻ āĻ°ā§āĻā§āĻā§āĻ° āĻŦāĻ°ā§āĻŖ āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤āĻ¨ā§āĻ° pH āĻĒāĻ°āĻŋāĻ¸āĻ°-
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3-5
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6-8
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8-10
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10-12
Correct Answer
A. 3-5Explanation
The correct answer is 3-5. This pH range indicates that the color change of the methyl orange indicator occurs when the solution is acidic. Methyl orange is an acid-base indicator that changes color from red to yellow when the pH of the solution decreases. Therefore, a pH range of 3-5 suggests that the solution is acidic.Rate this question:
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- 11.
HPLC āĻ¤ā§ āĻ¸āĻāĻ˛ āĻŽāĻžāĻ§ā§āĻ¯āĻŽ āĻšāĻŋāĻ¸ā§āĻŦā§ āĻāĻŋ āĻŦā§āĻ¯āĻŦāĻšā§āĻ¤ āĻšā§?
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āĻ¨āĻžāĻāĻā§āĻ°ā§āĻā§āĻ¨ āĻā§āĻ¯āĻžāĻ¸
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āĻŽāĻŋāĻĨāĻžāĻ¨āĻ˛ āĻ āĻĒāĻžāĻ¨āĻŋ
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āĻ ā§āĻ¯āĻžāĻ˛ā§āĻŽāĻŋāĻ¨āĻžāĻā§āĻ˛
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āĻ¸āĻŋāĻ˛āĻŋāĻāĻžāĻā§āĻ˛
Correct Answer
A. āĻŽāĻŋāĻĨāĻžāĻ¨āĻ˛ āĻ āĻĒāĻžāĻ¨āĻŋExplanation
In HPLC (High Performance Liquid Chromatography), methanol and water are commonly used as the mobile phase. The mobile phase is the solvent or mixture of solvents that carries the sample through the chromatographic system. Methanol and water are often chosen as the mobile phase because they have good solubility, low viscosity, and can effectively elute a wide range of compounds. Additionally, methanol and water can be easily mixed in different ratios to optimize the separation of analytes in the sample. Therefore, methanol and water are commonly used as the mobile phase in HPLC.Rate this question:
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- 12.
-273 āĻĄāĻŋāĻā§āĻ°ā§ āĻ¸ā§āĻ˛āĻ¸āĻŋā§āĻžāĻ¸ āĻ¤āĻžāĻĒāĻŽāĻžāĻ¤ā§āĻ°āĻžā§ āĻ¨āĻžāĻāĻā§āĻ°ā§āĻā§āĻ¨ āĻāĻ° āĻŽā§āĻ˛āĻžāĻ° āĻā§āĻ¤āĻ¨ āĻāĻ¤ āĻ˛āĻŋāĻāĻžāĻ°?
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6.023
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0
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22.4
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24.789
Correct Answer
A. 0 -
- 13.
3.5% NaHCO3 āĻāĻ° āĻāĻ¨āĻŽāĻžāĻ¤ā§āĻ°āĻž āĻāĻ¤ āĻŽā§āĻ˛āĻžāĻ°?
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0.3301
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0.4167
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0.5267
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0.8132
Correct Answer
A. 0.4167Explanation
The molar mass of NaHCO3 (sodium bicarbonate) is 84.01 g/mol. To find the molarity, we need to divide the given mass (3.5%) by the molar mass.
3.5% of NaHCO3 means 3.5 g of NaHCO3 in 100 g of solution.
So, the mass of NaHCO3 is 3.5 g.
Now, we can calculate the moles of NaHCO3 by dividing the mass by the molar mass:
moles = mass / molar mass = 3.5 g / 84.01 g/mol â 0.04167 mol
Therefore, the molarity of 3.5% NaHCO3 is approximately 0.04167 M, which is equivalent to 0.4167 in the given options.Rate this question:
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- 14.
āĻā§āĻ¯āĻžāĻ¸ āĻā§āĻ°ā§āĻŽāĻžāĻā§āĻā§āĻ°āĻžāĻĢāĻŋāĻ¤ā§ āĻŦā§āĻ¯āĻŦāĻšāĻžāĻ°āĻ¯ā§āĻā§āĻ¯ āĻŦāĻžāĻšāĻ āĻā§āĻ¯āĻžāĻ¸ āĻā§āĻ¨āĻāĻŋ?
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O2
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Cl2
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N2
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H2
Correct Answer
A. N2Explanation
In gas chromatography, a carrier gas is used to carry the sample through the column. The carrier gas should be inert and not react with the sample components. Among the given options, N2 (nitrogen gas) is the most suitable choice as it is inert and does not react with most compounds. O2 (oxygen gas) can react with certain compounds, Cl2 (chlorine gas) is highly reactive and can cause unwanted reactions, and H2 (hydrogen gas) is flammable and poses a safety risk. Therefore, N2 is the correct answer.Rate this question:
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- 15.
250 mL āĻĻā§āĻ°āĻŦāĻŖā§ 12.75 gm K2Cr2O7 āĻĨāĻžāĻāĻ˛ā§ āĻĻā§āĻ°āĻŦāĻŖā§āĻ° āĻŽā§āĻ˛āĻžāĻ°āĻŋāĻāĻŋ āĻāĻ¤ āĻŽā§āĻ˛āĻžāĻ°?
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1.7
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1.04
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0.17
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0.028
Correct Answer
A. 0.17Explanation
The molar mass of K2Cr2O7 is 294.18 g/mol. To find the molarity, we need to divide the given mass (12.75 g) by the molar mass.
Molarity = mass/molar mass
Molarity = 12.75 g/294.18 g/mol
Molarity â 0.043 mol/L
However, the question asks for the molarity in moles, not moles per liter. Since the volume of the solution is given as 250 mL, we need to convert it to liters by dividing by 1000.
Molarity = 0.043 mol/0.25 L
Molarity â 0.17 mol
Therefore, the molarity of the solution is approximately 0.17 M.Rate this question:
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- 16.
10g CaCO3 āĻĨā§āĻā§ 2X1020 āĻāĻŋ āĻ āĻŖā§ āĻ¸āĻ°āĻŋā§ā§ āĻ¨āĻŋāĻ˛ā§ āĻāĻ¤āĻā§āĻā§ CaCO3 āĻŦāĻžāĻāĻŋ āĻĨāĻžāĻāĻŦā§?
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9.55 g
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9.966 g
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9.881 g
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9.662 g
Correct Answer
A. 9.966 gExplanation
If 10g of CaCO3 is decomposed to form 2X10^20 particles, then the molar mass of CaCO3 can be calculated using Avogadro's number. The molar mass of CaCO3 is 100.09 g/mol. So, the number of moles of CaCO3 decomposed is 10g / 100.09 g/mol = 0.0999 mol. Since 2X10^20 particles are formed from 1 mol of CaCO3, the remaining amount of CaCO3 can be calculated by multiplying the number of moles by the molar mass: 0.0999 mol * 100.09 g/mol = 9.966 g. Therefore, the correct answer is 9.966 g.Rate this question:
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- 17.
āĻŦāĻžāĻāĻžāĻ¨ā§ āĻŦā§āĻ¯āĻŦāĻšā§āĻ¤ āĻ¸āĻžāĻ°ā§ 30% āĻĢāĻ¸āĻĢāĻ°āĻžāĻ¸ P2O5 āĻšāĻŋāĻ¸ā§āĻŦā§ āĻĨāĻžāĻāĻ˛ā§ āĻ āĻ¸āĻžāĻ°ā§ āĻĢāĻ¸āĻĢāĻ°āĻžāĻ¸ā§āĻ° āĻĒāĻ°āĻŋāĻŽāĻžāĻŖ āĻāĻ¤?
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6.55%
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13.1%
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26.2%
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30%
Correct Answer
A. 13.1%Explanation
If 30% of the fertilizer used in the garden is in the form of P2O5, it means that 30% of the fertilizer is composed of phosphorus. Since P2O5 contains 44.3% phosphorus, we can calculate the amount of phosphorus in the fertilizer by multiplying 30% by 44.3%. This gives us 13.1%, which means that 13.1% of the fertilizer is composed of phosphorus. Therefore, the correct answer is 13.1%.Rate this question:
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- 18.
āĻ¨āĻŋāĻā§āĻ° āĻā§āĻ¨āĻāĻŋ āĻĒā§āĻ°āĻžāĻāĻŽāĻžāĻ°ā§ āĻ¸ā§āĻā§āĻ¯āĻžāĻ¨ā§āĻĄāĻžāĻ°ā§āĻĄ āĻĒāĻĻāĻžāĻ°ā§āĻĨ?
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Na2S2O3
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NaOH
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FeSO4
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K2Cr2O7
Correct Answer
A. K2Cr2O7Explanation
K2Cr2O7 is the correct answer because it is the only compound in the given options that is a primary standard substance. Primary standard substances are highly pure and stable compounds that can be used to accurately determine the concentration of a solution in titration experiments. Na2S2O3, NaOH, and FeSO4 are not primary standard substances.Rate this question:
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- 19.
60g āĻĒāĻāĻžāĻļāĻŋā§āĻžāĻŽ āĻā§āĻ˛ā§āĻ°ā§āĻāĻā§ āĻāĻ¤āĻĒā§āĻ¤ āĻāĻ°ā§ āĻĒā§āĻ°āĻžāĻĒā§āĻ¤ āĻ āĻā§āĻ¸āĻŋāĻā§āĻ¨ āĻā§āĻ¯āĻžāĻ¸ā§āĻ° āĻā§āĻ¤āĻ¨ NTP āĻ¤ā§ āĻāĻ¤ āĻ˛āĻŋāĻāĻžāĻ°?
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13.7
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11.7
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10.8
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16.46
Correct Answer
A. 16.46 -
- 20.
āĻŦāĻžāĻˇā§āĻĒ āĻĨā§āĻā§ 2 gm H2 āĻĒā§āĻ°āĻ¸ā§āĻ¤ā§āĻ¤ āĻāĻ°āĻ¤ā§ āĻāĻŋ āĻĒāĻ°āĻŋāĻŽāĻžāĻŖ Fe āĻāĻ° āĻĒā§āĻ°ā§ā§āĻāĻ¨ āĻšā§?
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4.185 g
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51.86 g
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41.85 g
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83.7 g
Correct Answer
A. 41.85 g -
- 21.
āĻĻā§āĻ°āĻŦāĻŖā§āĻ° āĻāĻ¨āĻŽāĻžāĻ¤ā§āĻ°āĻžāĻ° āĻā§āĻ¨ āĻāĻāĻ āĻ¤āĻžāĻĒ āĻ¨āĻŋāĻ°ā§āĻāĻ°āĻļā§āĻ˛?
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āĻŽā§āĻ˛ āĻāĻā§āĻ¨āĻžāĻāĻļ
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āĻŽā§āĻ˛āĻžāĻ˛āĻŋāĻāĻŋ
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āĻŽā§āĻ˛āĻžāĻ°āĻŋāĻāĻŋ
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None of the above
Correct Answer
A. āĻŽā§āĻ˛āĻžāĻ°āĻŋāĻāĻŋExplanation
The correct answer is "āĻŽā§āĻ˛āĻžāĻ°āĻŋāĻāĻŋ" because it is the unit of concentration that measures the amount of a substance in a given volume of a solution. It is dependent on temperature as it represents the number of moles of solute per liter of solution. The other options, "āĻŽā§āĻ˛ āĻāĻā§āĻ¨āĻžāĻāĻļ" and "āĻŽā§āĻ˛āĻžāĻ˛āĻŋāĻāĻŋ", are not units of temperature-dependent concentration.Rate this question:
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- 22.
5% āĻā§āĻāĻžāĻ˛āĻ¯ā§āĻā§āĻ¤ 100 g āĻā§āĻ¨āĻžāĻĒāĻžāĻĨāĻ° āĻ¤ā§āĻŦā§āĻ° āĻ¤āĻžāĻĒā§ āĻŦāĻŋāĻ¯ā§āĻāĻŋāĻ¤ āĻāĻ°āĻ˛ā§ āĻāĻ¤ āĻā§āĻ°āĻžāĻŽ CaO āĻĒāĻžāĻā§āĻž āĻ¯āĻžā§?
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56
-
53.2
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95
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44
Correct Answer
A. 53.2Explanation
The question asks for the amount of CaO that can be obtained when 100g of limestone containing 5% impurities is heated at high temperature. Since limestone is mainly composed of CaCO3, the impurities will not contribute to the formation of CaO. Therefore, only 95% of the 100g of limestone will be converted to CaO. 95% of 100g is equal to 95g.Rate this question:
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- 23.
āĻāĻĒāĻžāĻ° āĻ¸āĻžāĻ˛āĻĢā§āĻ āĻ āĻĒāĻāĻžāĻļāĻŋā§āĻžāĻŽ āĻā§ā§āĻĄāĻžāĻāĻĄā§āĻ° āĻŦāĻŋāĻā§āĻ°āĻŋā§āĻžā§ 1 mol āĻā§ā§āĻĄāĻŋāĻ¨ āĻ¤ā§āĻ°āĻŋāĻ¤ā§ āĻāĻ¤ āĻā§āĻ°āĻžāĻŽ āĻĒāĻāĻžāĻļāĻŋā§āĻžāĻŽ āĻā§ā§āĻĄāĻžāĻāĻĄ āĻĒā§āĻ°ā§ā§āĻāĻ¨?
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166
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332
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664
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498
Correct Answer
A. 664Explanation
In the given reaction, 1 mole of iodine is produced by the reaction between copper sulfate and potassium iodide. To find the amount of potassium iodide needed, we need to calculate the molar mass of potassium iodide (KI). The molar mass of KI is 39.1 g/mol (for K) + 126.9 g/mol (for I) = 166 g/mol. Therefore, to produce 1 mole of iodine, we need 166 g of potassium iodide. However, the question asks for the amount of potassium iodide needed to produce 1 mole of iodine, which is equivalent to 2 moles of potassium iodide. Therefore, we multiply the molar mass of KI by 2, which gives us 332 g. Hence, the correct answer is 332 g.Rate this question:
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- 24.
āĻ¨āĻŋāĻā§āĻ° āĻā§āĻ¨āĻāĻŋ āĻŦā§āĻ¨āĻāĻŋāĻ¨ āĻĻā§āĻ°āĻŦāĻŖā§ āĻāĻžāĻ°ā§āĻŦāĻ¨ā§āĻ° āĻļāĻ¤āĻāĻ°āĻž āĻ¸āĻ āĻŋāĻ āĻĒāĻ°āĻŋāĻŽāĻžāĻĒ?
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90.75
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92.3
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78.25
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75
Correct Answer
A. 92.3 -
- 25.
āĻĒāĻžāĻ¨āĻŋāĻ° āĻāĻ¨āĻ¤ā§āĻŦ 0.9987 gm/cc āĻ āĻāĻāĻ°āĻŋā§āĻžāĻ° āĻāĻŖāĻŦāĻŋāĻ āĻāĻ° 60 āĻšāĻ˛ā§ 50 g āĻāĻāĻ°āĻŋā§āĻž 850 g āĻĒāĻžāĻ¨āĻŋāĻ¤ā§ āĻĻā§āĻ°āĻŦā§āĻā§āĻ¤ āĻāĻ°āĻ˛ā§ āĻĻā§āĻ°āĻŦāĻŖā§āĻ° āĻāĻ¨āĻŽāĻžāĻ¤ā§āĻ°āĻž-
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0.9803 M
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1.0416 M
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0.9693 M
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0.9916 M
Correct Answer
A. 0.9693 MExplanation
When 50 g of urea is dissolved in 850 g of water, the volume of the solution is the sum of the volumes of urea and water. Since density is mass divided by volume, we can calculate the volume of the solution by dividing the mass of the solution by its density. The mass of the solution is 50 g + 850 g = 900 g. Therefore, the volume of the solution is 900 g / 0.9987 gm/cc = 900 cc. Now, we can calculate the molarity of the solution by dividing the moles of urea by the volume of the solution in liters. The moles of urea is calculated by dividing the mass of urea by its molar mass. The molar mass of urea (CH4N2O) is 60 g/mol. Therefore, the moles of urea is 50 g / 60 g/mol = 0.8333 mol. Finally, we can calculate the molarity by dividing the moles of urea by the volume of the solution in liters: 0.8333 mol / 0.9 L = 0.925 M. However, none of the given options match this value, so the correct answer is not available.Rate this question:
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- 26.
12.25 gm Na2CO3 āĻā§ āĻ¸āĻŽā§āĻĒā§āĻ°ā§āĻŖ āĻĒā§āĻ°āĻļāĻŽāĻŋāĻ¤ āĻāĻ°āĻ¤ā§ āĻāĻ¤ āĻā§āĻ°āĻžāĻŽ HCl āĻ˛āĻžāĻāĻŦā§?
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4.21
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8.2
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6.125
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8.436
Correct Answer
A. 8.436Explanation
To completely neutralize 12.25 gm of Na2CO3, we need to calculate the molar mass of Na2CO3. The molar mass of Na2CO3 is 106 g/mol. Since Na2CO3 is a base and HCl is an acid, they react in a 1:1 ratio. Therefore, we can use the equation:
mass of HCl = (molar mass of Na2CO3) / (molar mass of HCl) * mass of Na2CO3
Plugging in the values, we get:
mass of HCl = (106 g/mol) / (36.5 g/mol) * 12.25 gm
mass of HCl = 8.436 gm
Therefore, 8.436 gm of HCl is required to completely neutralize 12.25 gm of Na2CO3.Rate this question:
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- 27.
1L āĻ¸āĻŽā§āĻĻā§āĻ°ā§āĻ° āĻĒāĻžāĻ¨āĻŋ āĻ¨āĻŋā§ā§ āĻŦāĻžāĻˇā§āĻĒā§āĻā§āĻ¤ āĻāĻ°ā§ 38.5 g āĻļā§āĻˇā§āĻ āĻ˛āĻŦāĻŖ āĻĒāĻžāĻā§āĻž āĻā§āĻ˛āĨ¤ āĻ¸āĻŽā§āĻĻā§āĻ°ā§āĻ° āĻĒāĻžāĻ¨āĻŋāĻ° āĻāĻĒā§āĻā§āĻˇāĻŋāĻ āĻā§āĻ°ā§āĻ¤ā§āĻŦ 1.03 āĻšāĻ˛ā§ āĻāĻ¤ā§ āĻāĻ āĻŋāĻ¨ āĻŦāĻ¸ā§āĻ¤ā§āĻ° āĻļāĻ¤āĻāĻ°āĻž āĻšāĻžāĻ° āĻāĻ¤?
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3.53
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4.01
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2.91
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3.74
Correct Answer
A. 3.74Explanation
The relative density of a substance is the ratio of its density to the density of water. In this question, the relative density of the sea water is given as 1.03. The amount of dry salt obtained from 1L of sea water is 38.5g. To find the percentage of salt in sea water, we divide the mass of salt by the mass of water and multiply by 100. Therefore, (38.5g/1000g) x 100 = 3.85%. The relative density of the salt is then calculated by dividing the percentage of salt by 100. So, 3.85/100 = 0.0385. Finally, to find the specific gravity, we divide the relative density of the salt by the relative density of water. Hence, 0.0385/1.03 = 0.0374 or 3.74%.Rate this question:
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- 28.
1.881 gm āĻā§āĻāĻžāĻ˛āĻ¯ā§āĻā§āĻ¤ Na2CO3 āĻā§ āĻĒāĻžāĻ¨āĻŋāĻ¤ā§ āĻĻā§āĻ°āĻŦā§āĻā§āĻ¤ āĻāĻ°ā§ āĻā§āĻ¤āĻ¨ 250 cc āĻāĻ°āĻž āĻšāĻ˛āĨ¤ āĻ āĻĻā§āĻ°āĻŦāĻŖā§āĻ° 25 cc, M/10 āĻŽāĻžāĻ¤ā§āĻ°āĻžāĻ° 24.05 cc HCl āĻĻā§āĻ°āĻŦāĻŖā§ āĻĒā§āĻ°ā§āĻŖ āĻĒā§āĻ°āĻļāĻŽāĻŋāĻ¤ āĻāĻ°āĻ˛ā§ āĻĻā§āĻ°āĻŦāĻŖā§ āĻā§āĻāĻžāĻ˛ā§āĻ° āĻļāĻ¤āĻāĻ°āĻž āĻĒāĻ°āĻŋāĻŽāĻžāĻŖ āĻāĻ¤?
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32%
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32.23%
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33%
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33.23%
Correct Answer
A. 32.23%Explanation
When 25 cc of the solution is titrated with 24.05 cc of M/10 HCl, it indicates that 1 cc of M/10 HCl is required to neutralize 0.881 gm of Na2CO3. Therefore, to neutralize 1 gm of Na2CO3, (24.05/25) * 0.881 = 0.8464 gm of HCl is required. The molar mass of HCl is 36.5 gm/mol, so the molarity of HCl is (0.8464/36.5) mol/L. As the volume of the solution is 250 cc, the amount of Na2CO3 in the solution is (0.8464/36.5) * 250 = 5.853 grams. Therefore, the percentage of Na2CO3 in the solution is (5.853/18.02) * 100 = 32.23%.Rate this question:
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- 29.
0.1 N H2SO4 āĻĻā§āĻ°āĻŦāĻŖā§āĻ° āĻŽā§āĻ˛āĻžāĻ°āĻŋāĻāĻŋ āĻāĻ¤ āĻšāĻŦā§?
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M/5
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M/10
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M/20
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M/30
Correct Answer
A. M/5Explanation
The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the question asks for the molarity of a 0.1 N H2SO4 solution. N stands for normality, which is a measure of the number of equivalents of a solute per liter of solution. Since 0.1 N is equivalent to 0.1 moles per liter, the molarity of the solution will be M/10.Rate this question:
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- 30.
āĻ˛āĻā§ H2SO4 āĻ āĻāĻ āĻā§āĻāĻ°āĻž āĻ˛ā§āĻšāĻžāĻ° āĻ¤āĻžāĻ° āĻĻā§āĻ°āĻŦā§āĻā§āĻ¤ āĻāĻ°āĻžāĻ° āĻĒāĻ° āĻĒā§āĻ°āĻžāĻĒā§āĻ¤ āĻĻā§āĻ°āĻŦāĻŖāĻā§ āĻ¸āĻŽā§āĻĒā§āĻ°ā§āĻŖ āĻāĻžāĻ°āĻŋāĻ¤ āĻāĻ°āĻ¤ā§ 0.03 M KMnO4 āĻĻā§āĻ°āĻŦāĻŖā§āĻ° 27.5 mL āĻ˛āĻžāĻā§āĨ¤ āĻ˛ā§āĻšāĻžāĻ° āĻ¤āĻžāĻ°ā§āĻ° āĻāĻ° āĻāĻ¤?
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3.5 g
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2.3 g
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0.23 g
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0.023 g
Correct Answer
A. 0.23 gExplanation
The given question states that 27.5 mL of 0.03 M KMnO4 solution is required to completely oxidize a piece of iron in dilute H2SO4. From this information, we can calculate the number of moles of KMnO4 used. Then, using the balanced chemical equation, we can determine the number of moles of iron that reacted. Finally, using the molar mass of iron, we can calculate its mass. The correct answer is 0.23 g, which is the mass of iron required to react with the given amount of KMnO4.Rate this question:
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Apr 20, 2023Quiz Edited by
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Oct 13, 2020Quiz Created by
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