Saturday, February 25, 2017

CBSE Class 9 Chemistry - Questions based on Mole Concept (#cbsenotes)(#class9SA2Chemistry)

Questions Based on Mole Concept

CBSE Class 9 Chemistry
CBSE Class 9 Chemistry - Questions based on Mole Concept (#cbsenotes)(#class9SA2Chemistry)

Q1: Fill in the blank.

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly ________ of C₁₂


Answer: 12g or 12 grams.



Q2: I owe you 1 mole of rupees. How many rupees do I have to pay you?


Answer: 6.022 × 10²³ rupees.


Q3: Is mole a unit?

Answer: No, it is a large constant number called Avagadro's constant (N) = 6.022 × 10²³


Q4: Atomic mass of Na is 23u. What is the mass of 1 mole of Na atoms? Is there any name for this?

Answer: 23g. It is also called molar mass of Na atoms.



Q5: Atomic mass of oxygen(O) is 16u. What's its molar mass?

Answer: 16g.



Q6: What is the molecular mass of Water molecule (H₂O)?

Answer: Atomic mass of H = 1u
Atomic mass of O = 16u

Molecular mass of H₂O = mass of 2 atoms of H + mass of 1 atom of O
= 2 × 1u + 16u
= 18u



Q7: What is the molar or gram molecular mass of water?

Answer: Molecular mass of Water (H₂O) = 18u
Gram molecular mass = 18g or 18 grams



Q8: How many moles will be in 45g of water(H₂O)?

Answer: Molecular mass of H₂O = 18u
Mass of 1 mole of H₂O = 18g
No. of moles in 45g of H₂O = 45 ÷ 18 = 2.5 moles



Q9: How many atoms of Hydrogen are there in 2 moles of water?

Answer: 1 molecule of H₂O has H-atoms = 2 atoms
1 mole of H₂O has H-atoms = 2 mole atoms
2 moles of H₂O has H-atoms = 2 × 2 = 4 moles of atoms
= 4 × 6.022 × 10²³
= 24.088 × 10²³
= 2.4088 × 10²⁴ H-atoms


Q10: How many grams of SiO₂ are present in 0.8 mol ? (molecular mass of SiO₂ = 60.1u)

Answer: Molecular mass of SiO₂ = 60.1u
1 mole of SiO₂ = 60.8g
0.8 mol of SiO₂ = 60.8 × 0.8 = 48.1g



Q11: Under standard temperature pressure conditions, what is the volume of 1 mol of gas?

Answer: 22.4 litres or 22.4l


Q12: How many mol of CO₂ (carbon dioxide) are present in 55.5 L ?

Answer: 1 mol of CO₂ = 22.4 l
55.5 litres contains = 55.5/22.4
= 2.48 mol of CO₂


Q13: What will be the mass of 12.044 × 10²³ molecules of O₂?

Answer: Molecular mass of O₂ = 2 × 16u = 32u
1 mol of O₂ = 32g
6.022 × 10²³ O₂ molecules have mass = 32g
  12.044 × 10²³ molecules of O₂ = 32 × 12.044 × 10²³ / (6.022 × 10²³)
  = 32 × 2
  = 64g


Q14: Calculate the mass of glucose in 2 × 10²⁴ molecules. Molecular mass glucouse is 180u

Answer: Molecular mass of glucouse = 180u
Gram molecular mass of glucouse = 180g
⇒ 1 mole (6.022 × 10²³) molecules = 180g
2 × 10²⁴ molecules will have mass = 180 × 2 × 10²⁴ / (6.022 × 10²³)
= 180 × 2 × 10 / 6.022 = 597.7 g


Q15: Write formula to calculate Number of moles 

(a) of monoatomic substance, if its mass is given
(b) of a molecular substance, if its mass is given
(c) of a monoatomic substance, if number of atoms are given
(d) of a molecular substance, if number of molecules are given


Answer:
(a)

No. of moles = mass
atomic mass

(b)
No. of moles = mass
molecular mass

(c)
No. of moles = no. of atoms
6.022 × 10²³

(d)
No. of moles = no. of molecules
6.022 × 10²³


Friday, February 24, 2017

CBSE Class 9 - कक्षा ९ - Hindi(B) हिंदी (ब) - कविता - नए इलाके में (प्रश्नोत्तर )

कविता - नए इलाके में 

कक्षा ९ - Hindi(B) हिंदी (ब) 
CBSE Class 9 - कक्षा ९ - Hindi(B) हिंदी (ब) - कविता - नए इलाके में (प्रश्नोत्तर )

प्रश्नोत्तर 


प्र १: नए बसते इलाके में कवि रास्ता क्यों भूल जाता है ?

उत्तर : नव निर्माण ही  कवि का रास्ता भूलना मुख्य कारण है । इसके अतिरिक्त अन्य कारण है :
१. नित्य नए मकान बनना
२. नए इलाकों का बसना
३. बनाए गए निशानों का न मिलना
४. नए परिवर्तन से कवि का दिशा भ्रमित हो जाना


प्र २ : कविता में कौन कौन से पुराने निशानों का उल्लेख किया गया है ?

उत्तर : कविता में निम्नलिखित पुराने निशानों का उल्लेख किया गया है :
(क) पीपल का पुराना  पेड़
(ख) ढहा हुआ मकान
(ग) खाली जंमीन का टुकड़ा
(घ) बिना रंग वाले लोहे के फाटक का एक मंजिला घर


प्र : 'नए इलाके में' कविता के रचियता का नाम लिखिए ?

उत्तर : अरुण कमल

Thursday, February 23, 2017

CHAPTER REVISION TEST - CLASS XI - PHYSICS (Work, Energy and Power) (#cbsenotes)

Work, Energy and Power

CHAPTER REVISION TEST  - CLASS XI - PHYSICS (Work, Energy and Power) (#cbsenotes)
CHAPTER REVISION TEST 
CLASS XI - PHYSICS 


Q1: When an air bubble rises in water, what happens to its potential energy? [1]

Q2: A spring is cut into two equal halves. How is the spring constant of each half affected? [1]

Q3: What should be the angle between the force and the displacement for maximum and minimum work? [1]

Q4: The momentum of an object is doubled. How does it’s. K.E. change? [1]

Q5: What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?  [1]

CBSE Class 10 - हिंदी (ब)/ Hindi(B) Marking Scheme (नमूना उत्तर पत्र ) of SA2 Sample Question Paper (2016-17) (#cbsepapers)

CBSE Class 10 - हिंदी (ब) / Hindi(B) Sample Question Paper (2016-17)

CBSE Class 10 - हिंदी (ब)/ Hindi(B) Marking Scheme (नमूना उत्तर पत्र ) of SA2 Sample Question Paper (2016-17)


See Hindi (B) SA2 Sample Question Paper (2016-17)

Wednesday, February 22, 2017

CBSE Class 10 - हिंदी (ब) / Hindi(B) Sample Question Paper (2016-17)

CBSE Class 10 - हिंदी (ब) / Hindi(B) Sample Question Paper (2016-17)

CBSE Class 10 - हिंदी (ब)/ Hindi(B) Sample Question Paper (2016-17)

See Hindi (B) SA2 Marking Scheme Of Sample Question Paper (2016-17)

CBSE Class 10 - NTSE Stage-1 Language Test-2 (#ntse)(#englishquiz)(#cbsenotes)

NTSE Stage-1 Language Test-2

CBSE Class 10 - NTSE Stage-1 Language Test-2

Direction: in question Nos. 1 – 6, Choose the alternative which expresses the meaning of the given idioms/phrases.


Q1: Once in a blue moon

(a) every month
(b) always
(c) rarely
(d) after mid night



Q2: Keep body and soul together

(a) to maintain life
(b) to be free from disease
(c) to live in a joint family
(d) to have fun in life




Tuesday, February 21, 2017

GGS INDRAPRASTHA UNIVERSITY Common Entrance Test for Graduate Courses (2017-18)

GGS INDRAPRASTHA UNIVERSITY, DELHI Applications Invited for Common Entrance Test for Graduate Courses

GGS INDRAPRASTHA UNIVERSITY Common Entrance Test for Graduate Courses (2017-18)

Guru Gobind Singh Indraprastha University, Delhi has commenced Admission process for 2017-18. You may apply for Common Entrance Test (CET) for various graduate courses.

Important Courses are listed below:

Sl.No CET Name CET Code Last Date of Application (UPTO 4 PM) CET Date CET Time Date of Declaration of Result (Latest by)
1. BTECH 131 4th April, 2017 Tuesday 23rd April, 2017 Sunday 2.00 PM to 4.30 PM 9th May, 2017 Tuesday
2. BJMC 126 4th April, 2017 Tuesday 29th April, 2017 Saturday 2 PM to 4.30 PM 16th May, 2017 Tuesday