Bihar STET Paper 2 Chemistry Mock Test - 8 - Bihar PGT/TGT/PRT MCQ

The correct answer is option  D.
Milk is an emulsion with fat particles (globules) dispersed in an aqueous (watery) environment. The fat globules do not coalesce and form a separate layer (oil off or churn) because they are protected by a membrane layer which keeps the fat particles separate from the water phase.

Rate constant is independent of concentration of reactant.

Enzymes are biological catalyst, they catalize different chemical reactions in body.

= 2.53/106   2.59 × 10^-2  5.9 × 10^-2
0.238 × 0.25          = 1.18 × 10^-3

250 ml - 0.238

= 0.955

0.955  0.955 

H₂O₂ is an oxidizing agent in 1st reaction and Reducing agent in 2nd reaction.

Recycling of materials and energy will lead to decrease in environmental pollution

Constant Temperature means isotherm.

Dobereiner's Triads were groups of three elements with similar chemical properties, where the atomic mass of the middle element was approximately equal to the average of the other two. Carbon, being a nonmetal with a unique set of properties, did not fit into any known triad.

The basicity decreases from Ce(OH)₃ to Lu(OH)₃. Due to lanthanide contraction, the decrease in the size of the cation increases the covalent character between the lanthanide ion and hydroxide ion thereby, reducing the basic character of the lanthanide hydroxide.

 NH4HS (s)  ⇋ NH3 (g) + H2S (g)
Initial    1                   -               -
At eqm     1-x                 x+0.02     x
Kc = [NH₃][H₂S]   (Since NH4HS is solid, we ignore it.)
1.8×10-4    = (x+0.02)(x)
x2+0.02x-1.8×10-4 = 0
Applying quadratic formula; x = -0.02+√{(0.02)2-4×1.8×10-4}
= 0.033-0.020/2 = 0.0065
Therefore, concn of NH₃ at equilibrium = x+0.020 = 0.0265
concn of H2S at equilibrium = x = 0.0065
So, option b is correct

As per IUPAC nomenclature, the roots are as follows: 0 – nil, 1 – un, 2 – bi, 3 – tri, 4 – quad, 5 – pent, 6 – hex, 7 – sept, 8 – oct and 9 – enn. So the roots for 1, 1 and 9 are un, un, and enn. Hence the symbol of the elements is Uue.

The correct answer is option B
When conc. of B is doubled, the rate is doubled. 
So, Order w.r.t.  [B] is 1.
The rate increases by a factor of 8 when conc. of [A] and [B] are doubled.
So, Order w.r.t.  [A] is 2.
Overall Rate is 
Rate=k[A]²[B]

The complex [CO(NH₃)₅NO₂]CI₂  produces three ions in aqueous solution.

[Co(NH₃)₅NO₂]CI₂ → [Co(NH₃)₅NO₂]⁺ + 2Cl^-
The aqueous solution on treatment with an excess of AgNO₃ gives two moles of AgCI because we have 2 moles of Cl^- ion in aqueous solution.
Linkage isomers

[Co(NH₃)₅NO₂]CI₂ and [Co(NH₃)₅(ONO)]CI₂
The nitrite ligand is bound through oxygen (—ONO)in red form. The nitrite ligand is bound through nitrogen (—NO₂) in yellow form.

O — H group is attached to (ene) is m ost a cidic and thus, has maximum intermolecular H-bonding.

T.S ∞ Ionic character (for polyatomic anion)
Li₂CO₃ is least ionic or most covalent

Structure with least formal charge on each atom, makes greater contributions. Also, structures with at least one neutral atom is also favoured II and III identical.

For the equilibrium reaction:
A+2B ⇌ 2C+D
volume of flask = 1L
Initial moles of A = 2 mol
initial concentration of A=[A]i = 2 M
initial mole of B = 1 mol 
[B]i = 1 M
[A]_eq = 2-x, [B]_eq = 1-2x, [C]_eq = x, [D]_eq = 3x
Given [D]_eq = 1 * 1L
= 1 M
Thus x = 1M
[A]_eq = 1, [B]_eq = -1, [C]_eq = 1, [D] = 3
Kc = {([D]eq)³ * ([C]eq)}/{[A]eq * ([B]eq)²} 
= Kc = {(3)^3*1}/{1*(-1)²}
= 27/1
= 27

The p-block elements constitute elements belonging to group 13 to 18 and these together with the s-Block elements are called Representative elements.

Teaching process includes three variables:

• Dependent variable: A student is a dependent variable.
• Independent variable: A teacher is an independent variable.
• Intervening variable: It plays a role for the interaction between other two variables. This includes content and methods of teaching.

Teachers use different evaluation methods because different subjects require different methods of assessment like science can be assessed by practical exams and languages can be assessed by evaluating verbal ability and fluency.

Disciplinary issues come up when a teacher fails to engage students in a meaningful and constructive programme. The nature of students is volatile. They are easily distracted by things. Only a constructive programme could help them gain attention towards the meaningful tasks conducted in the classroom, so that their mind remains occupied to the most productive tasks.

The teacher is seen as a facilitator of learning. The progress of learning varies from individual to individual. Knowledge and values are reasoned and inquiry is the preferred mode of learning.

Key Points

•  A teacher’s duty is to help students develop intellectually, socially, physically, emotionally, and morally so as to bring about their all-round development. 
• Teachers should be learning facilitators rather than didactic instructors or preachers and in order to be successful as learning facilitators, they should be trained to be sensitive and caring, genuine and empathetic.
•  A good teacher must be resourceful and participative.
• Effective teachers emphasize healthy social and interpersonal development and test-oriented approaches to the subject matter. 
• Effective teachers are found to have more elaborate systems of knowledge for analyzing problems in teaching. 
• Content mastery or deep knowledge of the subject that the teacher is to teach is an important trait of an ideal teacher.
• Teachers who are firmly rooted in their subject knowledge make clearer presentations and recognize students' difficulties readily. They can handle the queries of their students effectively

Thus, it is concluded that a good teacher must be resourceful and participative.

An EVS teacher should accept and respect perception of children. She must respect and accept the students and their perceptions.

The figure is given below:-

Sunny faces east direction then move 45° clockwise direction and the move 135° anticlockwise direction. Now Sunny faces north direction.

Hence, "option 2" is the correct answer.

1) GRASS – FEARLESS (Cannot be formed because G is missing)

2) RESEAL – FEARLESS (Can be formed)

3) LESSER – FEARLESS (Can be formed)

4) ERASE – FEARLESS (Can be formed)

Hence, the correct answer is “GRASS”.