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प्रश्न
How does calcination differ from roasting?
उत्तर
| Roasting | Calcination |
| Ore is heated in excess of air. | Ore is heated in the absence or limited supply of air. |
| This is used for sulphide ores. | This is used for carbonate ores. |
| SO2 is produced along with metal oxide. | CO2 is produced along with metal oxide. |
| e.g. \[\ce{ 2ZnS + 3O2 ->[\Delta] 2ZnO + 2SO2}\] | e.g. \[\ce{ZnCO3 ->[\Delta] ZnO + CO2}\] |
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संबंधित प्रश्न
In a first order reaction x → y, 40% of the given sample of compound remains unreacted in 45 minutes. Calculate rate constant of the reaction.
A → B is a first order reaction with rate 6.6 × 10-5m-s-1. When [A] is 0.6m, rate constant of the reaction is
- 1.1 × 10-5s-1
- 1.1 × 10-4s-1
- 9 × 10-5s-1
- 9 × 10-4s-1
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Write molecularity of the following reaction:
2NO(g)+O2(g)→2NO2(g)
For the first order thermal decomposition reaction, the following data were obtained:
Time / sec Totalpressure / atm
0 0.30
300 0.50
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| t/s | 0 | 30 | 60 |
| [CH3COOCH3] / mol L–1 | 0.60 | 0.30 | 0.15 |
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
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For a reaction, \[\ce{A + B -> Product}\]; the rate law is given by, `r = k[A]^(1/2)[B]^2`. What is the order of the reaction?
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\[\ce{CH3CHO_{(g)} -> CH4_{(g)} + CO_{(g)}}\] Rate = k [CH3CHO]3/2
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| A/mol L−1 | 0.20 | 0.20 | 0.40 |
| B/mol L−1 | 0.30 | 0.10 | 0.05 |
| r0/mol L−1 s−1 | 5.07 × 10−5 | 5.07 × 10−5 | 1.43 × 10−4 |
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\[\ce{A(g) + 2B(g) -> 2C(g)}\]
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| Experiment | Initial concentration of [A]/mol L–¹ |
Initial concentration of [B]/mol L–¹ |
Initial rate of formation of [C]/mol L–¹ s–¹ |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
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Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.
| Column I | Column II | |
| (i) |  |
|
| (ii) |  |
(a) 1st order |
| (iii) |  |
(b) Zero-order |
| (iv) |  |
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Reason: We cannot determine order from balanced chemical equation.
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Reason: A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.
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[Given: [At] in M and t in sec.]
[Express your answer in terms of 10-2 M /s]
[Round off your answer if required]
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