👨🏻‍🔬 | 𝓟𝓱𝓪𝓻𝓶𝓪 𝓟𝓾𝓵𝓼𝓮 |👩🏻‍🔬

‏{رَبِّ اشْرَحْ لِي صَدْرِي ۝ وَيَسِّرْ لِي أَمْرِي}

هاي mcq نفس النقاط الي فوك بس بملف هنا 🤍

هاي شرح الموضوع كامل 🤍

68سؤال كلامي 12 سؤال مسائل وكلهن وزاريات 🤍🤝🏻

يدللون علينا العيالة 🤍

عاشت ايدك حجي واحد مرشحات

Answer: A Q36 Bulkiness equals reciprocal of: A) True density B) Bulk density C) Granule density D) Particle density Answer: B Q37 Bulkiness increases when particle size: A) Increases B) Decreases C) ثابت D) disappears Answer: B Q38 Flow properties affected by: A) Particle size B) Particle shape C) Surface texture D) All of the above Answer: D Q39 Particles smaller than 10 µm show: A) Excellent flow B) Poor flow C) No flow D) Perfect packing Answer: B Q40 Poor powder flow due to moisture can be improved by: A) Heating B) Drying C) Cooling D) Filtering Answer: B

Q1 Micromeritics is the science concerned with: A) Liquids B) Small particles C) Gases D) Solutions Answer: B Q2 1 micrometer equals: A) 10⁻³ m B) 10⁻⁶ m C) 10⁻⁹ m D) 10⁻² m Answer: B Q3 Polydisperse system means particles: A) Same shape B) Same density C) Different sizes D) Same surface area Answer: C Q4 Equivalent spherical diameter is used when particles are: A) Spherical B) Regular C) Irregular D) Uniform Answer: C Q5 Surface diameter (ds) is used in: A) Dissolution B) Adsorption C) Both A and B D) Density Answer: C Q6 Volume diameter (dv) relates to: A) Density B) Mass C) Volume D) All of the above Answer: D Q7 Stokes diameter depends on: A) Surface area B) Sedimentation rate C) Shape only D) Weight only Answer: B Q8 Particle size distribution curve plots particle size versus: A) Porosity B) Density C) Frequency D) Volume Answer: C Q9 Frequency distribution curve Y-axis represents: A) Particle size B) Particle diameter C) Frequency D) Volume Answer: C Q10 Optical microscope measures particles within range: A) 0.2–100 µm B) 1–1000 µm C) 44–500 µm D) 10–200 µm Answer: A Q11 Recommended particle number counted microscopically: A) 50–100 B) 100–200 C) 300–500 D) 500–1000 Answer: C Q12 Microscopy disadvantage: A) Expensive B) Measures only two dimensions C) Measures depth accurately D) Fast method Answer: B Q13 Sieving method suitable mainly for: A) Fine particles B) Coarse particles C) Nanoparticles D) Liquids Answer: B Q14 Smallest particle size measurable by sieving: A) 10 µm B) 25 µm C) 44 µm D) 100 µm Answer: C Q15 Sieving errors caused by: A) Sample weight B) Shaking time C) Agitation intensity D) All of the above Answer: D Q16 Sedimentation method applies to particles smaller than: A) 100 µm B) 44 µm C) 10 µm D) 1 µm Answer: B Q17 Sedimentation method is based on: A) Boyle’s law B) Stokes’ law C) Newton’s law D) Raoult’s law Answer: B Q18 Stokes law assumes particles: A) Aggregated B) Cubical C) Spherical D) Rectangular Answer: C Q19 Stokes law valid when particles move with: A) Variable velocity B) Constant velocity C) Zero velocity D) High velocity Answer: B Q20 Particle aggregation during sedimentation causes: A) Accurate results B) Faster settling C) Slower settling D) No change Answer: B Q21 Particle shape affects: A) Flow B) Packing C) Surface area D) All of the above Answer: D Q22 Spherical particles have: A) Maximum surface area B) Minimum surface area C) No surface area D) Infinite surface area Answer: B Q23 Irregular particles show: A) Better flow B) Loose packing C) Larger surface area D) Both B and C Answer: D Q24 Increase surface area increases: A) Density B) Solubility C) Weight D) Volume Answer: B Q25 Specific surface equals surface area per unit: A) Weight B) Volume C) Both D) Length Answer: C Q26 BET method determines: A) Porosity B) Density C) Surface area D) Diameter Answer: C Q27 Air permeability method depends on: A) Pressure only B) Surface area C) Temperature only D) Volume Answer: B Q28 Increase surface area leads to: A) Increased airflow B) Increased resistance C) No change D) Decreased adsorption Answer: B Q29 Porosity equals: A) Bulk volume / void volume B) Void volume / bulk volume C) True volume / bulk volume D) Weight / volume Answer: B Q30 Closest packing porosity: A) 26% B) 48% C) 30% D) 50% Answer: A Q31 Loosest packing porosity: A) 26% B) 48% C) 30% D) 60% Answer: B Q32 Most powders porosity ranges between: A) 10–20% B) 30–50% C) 50–70% D) 70–90% Answer: B Q33 True density excludes: A) Interparticle spaces B) Intraparticle pores C) Crystal lattice voids D) All of the above Answer: D Q34 Granule density measured using: A) Water B) Mercury C) Alcohol D) Oil Answer: B Q35 Bulk density calculated from: A) Weight and bulk volume B) Weight and true volume C) Density and volume D) Surface area

Q1 – Porosity A powder sample has: Bulk volume = 80 cm³ True volume = 60 cm³ Porosity equals: A) 10% B) 20% C) 25% D) 40% الحل الوزاري المختصر: Void volume = 80 − 60 = 20 Porosity = 20 / 80 = 0.25 = 25% Answer: C Q2 – Porosity Bulk volume = 100 cm³ Void volume = 30 cm³ Porosity equals: A) 0.3 B) 0.7 C) 3 D) 30 Answer: A (لأن Porosity = void / bulk) Q3 – True volume Weight = 120 g True density = 3 g/cm³ True volume equals: A) 20 B) 30 C) 40 D) 60 الحل: Volume = weight / density = 120 / 3 = 40 Answer: C Q4 – Bulk density Weight = 100 g Bulk volume = 50 cm³ Bulk density equals: A) 0.5 B) 2 C) 5 D) 10 الحل: Bulk density = weight / bulk volume = 100 / 50 = 2 Answer: B Q5 – Bulkiness Bulk density = 0.5 g/cm³ Bulkiness equals: A) 0.5 B) 1 C) 2 D) 5 الحل: Bulkiness = 1 / bulk density = 1 / 0.5 = 2 Answer: C Q6 – Void volume Bulk volume = 90 cm³ True volume = 70 cm³ Void volume equals: A) 10 B) 20 C) 30 D) 40 الحل: Void = bulk − true = 90 − 70 = 20 Answer: B Q7 – Porosity percent Void volume = 25 Bulk volume = 100 Porosity (%) equals: A) 15% B) 20% C) 25% D) 30% Answer: C Q8 – True volume from density Weight = 200 g True density = 4 g/cm³ True volume equals: A) 25 B) 50 C) 75 D) 100 الحل: Volume = weight / density = 200 / 4 = 50 Answer: B Q9 – Packing arrangement Porosity theoretical value in closest packing equals: A) 26% B) 30% C) 48% D) 50% Answer: A Q10 – Packing arrangement Porosity theoretical value in loosest packing equals: A) 26% B) 48% C) 30% D) 50% Answer: B Q11 – Bulk density calculation Weight = 150 g Bulk volume = 75 cm³ Bulk density equals: A) 1 B) 2 C) 3 D) 4 الحل: 150 / 75 = 2 Answer: B Q12 – Porosity calculation (نمط وزاري مشهور جدًا) Weight = 131.3 g True density = 3.203 g/cm³ Bulk volume = 82 cm³ True volume equals: A) 20 B) 30 C) 41 D) 60 الحل: True volume = weight / density ≈ 131.3 / 3.203 ≈ 41 Answer: C (هذا المثال نفسه موجود بالمحاضرة وغالبًا يجي امتحان) أهم القوانين الوزارية اللي تحفظها قبل الامتحان ⭐ 1️⃣ Porosity ε = Void volume / Bulk volume 2️⃣ Void volume Void = Bulk − True 3️⃣ True volume True volume = Weight / True density 4️⃣ Bulk density Bulk density = Weight / Bulk volume 5️⃣ Bulkiness Bulkiness = 1 / Bulk density

MCQ وزاري – Micromeritics Q1 The science that deals with small particles is called: A) Rheology B) Micromeritics C) Diffusion D) Sedimentation Answer: B Q2 The unit most commonly used in micromeritics is: A) nm B) cm C) µm D) mm Answer: C Q3 Surface diameter (ds) is based on: A) Volume B) Sedimentation rate C) Surface area D) Density Answer: C Q4 Volume diameter (dv) is used in determination of: A) Density and mass B) Flow rate C) Adsorption D) Dissolution Answer: A Q5 Stokes diameter (dst) depends on: A) Surface area B) Sedimentation velocity C) Particle color D) Shape only Answer: B Q6 Equivalent spherical diameter is used because: A) Particles are always spherical B) Particles are always cubic C) Irregular particles have no single true diameter D) Particles dissolve rapidly Answer: C Q7 Particle size distribution curve plots particle size versus: A) Density B) Frequency C) Shape D) Viscosity Answer: B Q8 In particle size distribution curve: X-axis represents: A) Frequency B) Density C) Particle size D) Porosity Answer: C Q9 Optical microscopy measures particle size range: A) 0.001 – 1 µm B) 0.2 – 100 µm C) 100 – 500 µm D) 1 – 1000 µm Answer: B Q10 Recommended number of particles counted in microscopy method: A) 10–50 B) 50–100 C) 300–500 D) 1000 Answer: C Q11 Main disadvantage of microscopic method: A) Expensive B) Measures only two dimensions C) Cannot measure size D) Requires heating Answer: B Q12 Sieving method is suitable mainly for: A) Fine particles B) Coarse particles C) Nanoparticles D) Liquids Answer: B Q13 Minimum particle size measurable by sieving: A) 10 µm B) 25 µm C) 44 µm D) 100 µm Answer: C Q14 Sedimentation method is based on: A) Newton’s law B) Stokes’ law C) Boyle’s law D) Fick’s law Answer: B Q15 Stokes’ law applies exactly when particles are: A) Cubic B) Aggregated C) Spherical D) Colored Answer: C Q16 Stokes’ equation is valid when particles: A) Are aggregated B) Move with constant velocity C) Are irregular D) Are heated Answer: B Q17 Particle shape affects: A) Flow properties B) Packing properties C) Surface area D) All of the above Answer: D Q18 Spherical particles have: A) Maximum surface area B) Minimum surface area C) No surface area D) Variable surface area Answer: B Q19 Increase in surface area leads to increase in: A) Density B) Solubility C) Viscosity D) Weight Answer: B Q20 BET method measures: A) Density B) Surface area C) Particle size D) Porosity Answer: B Q21 Air permeability method depends on: A) Surface area B) Color C) Density D) Weight Answer: A Q22 Porosity is defined as ratio of: A) True volume / bulk volume B) Void volume / bulk volume C) Bulk volume / void volume D) Weight / volume Answer: B Q23 Porosity of closest packing equals: A) 26% B) 48% C) 30% D) 50% Answer: A Q24 Porosity of loosest packing equals: A) 26% B) 48% C) 30% D) 60% Answer: B Q25 Bulk density depends mainly on: A) Particle size B) Particle shape C) Particle adhesion D) All of the above Answer: D Q26 Bulkiness is the reciprocal of: A) True density B) Granule density C) Bulk density D) Porosity Answer: C Q27 Particles smaller than 10 µm show poor flow due to: A) Gravity increase B) Cohesive forces C) Density decrease D) Surface tension Answer: B Q28 Flow rate becomes maximum at: A) Small particle size B) Medium particle size C) Very large particle size D) Nanoparticles Answer: B

هذا التقرير مال ماده فيزياء اذا احد يريد يكتب فهذا جاهز بدون ماتدور وتتعب

Research: Dissolving 1 Milliliter of Paracetamol in 10 Grams of Ethanol 1. Introduction Paracetamol (also known as acetaminophen) is one of the most widely used medications in the world. It is commonly used as an analgesic (pain reliever) and antipyretic (fever reducer). This compound has physical and chemical properties that allow it to dissolve in several organic solvents such as ethanol, while its solubility in water is relatively lower. Ethanol is a polar organic solvent widely used in chemistry and pharmaceutical sciences due to its ability to dissolve many organic compounds. Scientific studies indicate that the solubility of paracetamol in ethanol is approximately 18 grams per 100 mL at room temperature, which is significantly higher than its solubility in water. This experiment aims to study the dissolution process between paracetamol and ethanol and observe the formation of a homogeneous solution. 2. Aim of the Experiment The objectives of this experiment are: To study the solubility of paracetamol in ethanol. To observe the physical changes during the dissolution process. To understand the interaction between an organic solvent and a pharmaceutical compound. To prepare a homogeneous solution of paracetamol. 3. Materials and Apparatus Chemicals Paracetamol Ethanol Apparatus Electronic balance Graduated cylinder Beaker Glass stirring rod Pipette Thermometer (optional) 4. Theory Dissolution Dissolution is the process in which a solid substance (solute) mixes with a liquid substance (solvent) to form a homogeneous solution. The dissolution process depends on several factors: Nature of the solute and solvent Temperature Stirring Surface area of the solute Ethanol is a partially polar solvent capable of dissolving organic compounds like paracetamol due to hydrogen bonding interactions between molecules. Solubility The solubility of paracetamol in ethanol is relatively high and can reach approximately 18 g per 100 mL of ethanol at room temperature. Therefore, a small amount of paracetamol can dissolve easily in ethanol. 5. Experimental Procedure Measure approximately 1 milliliter of paracetamol (or the equivalent mass) using an electronic balance. Measure 10 grams of ethanol using a graduated cylinder. Pour the ethanol into a clean beaker. Gradually add the paracetamol to the ethanol. Stir the mixture using a glass rod. Observe the dissolution process until a clear solution forms. Record observations such as: Rate of dissolution Color of the solution Presence or absence of residue. 6. Expected Results After adding paracetamol to ethanol and stirring: The paracetamol gradually dissolves in the ethanol. A clear or slightly transparent solution forms. No solid residue should appear if the amount added is within the solubility limit. 7. Discussion The dissolution rate in this experiment depends on several factors: 1. Nature of the Solvent Ethanol is partially polar, which allows it to form hydrogen bonds with paracetamol molecules. 2. Stirring Stirring increases the rate of dissolution by improving contact between the solute and the solvent. 3. Temperature Higher temperature increases molecular motion, which accelerates the dissolution process. 4. Solution Concentration If the amount of paracetamol exceeds its solubility limit, some of the solid will remain undissolved and settle at the bottom of the container. 8. Safety Precautions Ethanol should be handled away from open flames because it is highly flammable. Wear gloves and safety goggles during the experiment. Work in a well-ventilated area. Avoid contact of chemicals with skin or eyes. 9. Conclusion The experiment demonstrates that paracetamol dissolves readily in ethanol due to interactions between the polar molecules of the solvent and the solute. The results confirm that ethanol is an effective solvent for preparing paracetamol solutions, which explains its use in certain pharmaceutical and chemical applications. In addition, factors such as stirring, temperature, and concentration significantly influence the efficiency and speed of the dissolution process

درجات بالضيم 🤣