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When sodium acetate on heating with soda lime, methane is produced. Note: In decarboxylation, the sodium salt of carboxylic acid on heating with soda lime produces alkane with one less carbon atom than a carboxylic acid. In decarboxylation, the product is alkane having one less carbon atom than a carboxylic acid.

Given:

The ratio of the quantities of acid and water in a mixture is 1 : 3

After added 5 L of acid, the new ratio becomes 1 : 2

Concept Used:

New quantity of Acid/New quantity of Water = (Quantity of Acid + Quantity of Added Acid)/(Quantity of Water + Quantity of Added Water)

Calculation:

Let the quantity of acid in the original mixture be x litres.

And, that of water be 3x L

New quantity of Acid/New quantity of Water = (Quantity of Acid + Quantity of Added Acid)/(Quantity of Water + Quantity of Added Water)

⇒ (x + 5)/3x = 1/2

⇒ 2x + 10 = 3x

⇒ x = 10

Quantity of new mixture = 4x + 5

⇒ 4 × 10 + 5 = 45 litres

∴  The quantity of new mixture in liters is 45 litres.

1. goes 

2. are gathered 

3. been, waiting 

4. is moving/moves 

5. attends 

6. enjoy / are enjoying 

7. is working 

8. is pouring

1. According to the teachers, Janu is studying in Class X.

2. In spite of her poverty, she is a disciplined student.

3. Because of her neatness, many students like her.

4. Apart from her studies, she is also known for her simplicity.

5. She gets on with everyone in spite of her participation in sports.

6. Due to her opposition to take leave, she ensures her age and gender in the school.

7. On account of completing the work given before, she goes to school the next day.

Causes of neurosis

According to Sigmund Freud (1856-1939), a famous Austrian neurologist who founded the discipline of psychoanalysis, neurosis is a coping strategy caused by unsuccessfully repressed emotions from past experiences.

These emotions overwhelm or interfere with current experience. He gave the example of an overwhelming fear of dogs that may have resulted from a dog-attack earlier in life.

Causes of Neurosis

In modern science, two components are considered among the causes of the development of neurosis: psychogenic and biological.

The inability to adequately respond to external stimuli, which causes stress, is one of the psychogenic causes of neurotic disorders. The first reason for an inadequate reaction could be a low-stress tolerance, as well as high susceptibility to even those things that do not cause a pathological reaction from the psyche in stronger people. Dripping water from a faucet, for example, can cause severe irritability in some people, while others simply do not notice these sounds. As a result, the former has a much higher proclivity and risk of developing a neurosis. It is impossible to discuss the presence of such a person in psychopathology because it is a personality trait, a character caused by a hereditary, genetic or acquired (as a result of upbringing or social environment) factor.

The second cause of psychogenic neurosis is a strong external stimulus that even people with a "strong psyche" find difficult to cope with. Chronic stress at work (professional failures, interpersonal relationships, management bias), conflicts at home, domestic problems, health problems, loss of loved ones, financial problems, and so on are examples of such irritants. Overwork, a lack of adequate rest, and an inability to relax all increase the risk of neurosis.

The biological cause of neurotic disorders is a disruption in the metabolism of neurotransmitters, hormones, vitamins, and other biologically active substances that are essential for the normal functioning of the central nervous system and, in particular, higher nervous activity. This reason cannot be considered separately from psychogenic neurosis because, in the end, it all comes down to impaired brain neuron function; however, what is the primary factor in pathological changes in nerve cells is considered to be the primary cause of neurotic disorder.

1. Be Organized
2. Communicate Effectively
3. Lead by Example
4. Enjoy Solving Problems
5. Be Comfortable Making Difficult Decisions

5. Strong base with a chemical formula NaOH. 

6. It is obtained from Lichens .

7.Use : To test whether a solution is basic or acidic .

8. Solutions which are neither basic or acidic are called neutral substances.

a W = mgh = 2 × 9.8 × 5 = + 98 Jb The work done by gravity is –98 J

In mechanics we treat length, mass and time as basic or fundamental quantities. Derived quantities : The quantities, which can be expressed in terms of the fundamental quantities are called derived quantities. For example volume, velocity, Pressure etc.

For a healthy person it is necessary that 

(a) The surrounding environment should be clean. Air and water borne diseases will not spread. 

(b) Personal hygiene prevents infectious diseases. 

(c) Proper, sufficient nourishment and food is necessary for good immune system of our body. 

(d) Immunisation against severe diseases.

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3 

Number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)

 = 3 × 3 = 9 

Thus, the number of elements in (A × B) is 9. 

Hint— Due to poor immune system, some children fall ill frequently. Balanced diet and proper nutrition for healthy body is required to have a strong immune system.

Let x = 0.2bar(54) = 0.2545454... … (1) 

100x = 25.45454545…. … (2) 

Now,

Subtracting equation (1) from equation (2), we get 

100x – x = (25.454545….) – (0.25454545….) 

⇒ 99x = 25.2 

= \(\frac{252}{10}\) 

= \(\frac{126}{5\times 99}\)

⇒ x = \(\frac{126}{5\times 99}\) 

= \(\frac{14}{5\times 11}\) 

= \(\frac{14}{55}\). 

Hence, 

Simplest form of \(0.2\overline{54}\) is \(\frac{14}{55}\).

Given that,

A(3, 0), B(4, 5), C(–1, 4)and D (–2, –1) are vertices of rhombus. 

We know that,

Area of rhombus = \(\frac{1}{2}\) × product of diagonals 

= \(\frac{1}{2}\) × AC × BD 

Now, 

AC = \(\sqrt{(−1 − 3)^2 + (4 − 0)^2}\) 

(By distance formula)

= \(\sqrt{16 + 16}\)

= 4√2. 

And BD= \(\sqrt{(−2 − 4) ^2 + (−1 − 5) ^2}\) 

= \(\sqrt{36 + 36}\) 

= 6√2. 

∴ Area of rhombus = \(\frac{1}{2}\) × AC × BD 

= \(\frac{1}{2}\) × 4√2 × 6√2 

= 24 square units.

Hence,

Area of rhombus = 24 square units.

f(x)= \(x^\frac12\)(x - 2), x ∈ R & x > 0 .

Domain of  f(x) is (0 , 0).

∴ f¹(x) = \(x^\frac12\) + \(\frac12\)\(x^\frac{-1}2\)(x - 2)

for critical point, we have f¹(x) = 0

∴ \(x^\frac12\) + \(\frac12\)\(x^\frac{-1}2\)(x - 2) = 0

\(\Rightarrow\) 2x + x - 2 = 0 ( on multiplying both sides by \(2x^\frac{1}2\))

\(\Rightarrow\) 3x - 2 = 0

\(\Rightarrow\) x = \(\frac23\)

Hence, x = \(\frac23\) is a critical point of f.

Given equation is 

3x³ - 2x² - x + 1 = 0----(1)

Let α, ß, γ  are root of equation (1)

\(\therefore\) Sum of roots = -b/a = -(-2)/3 = 2/3

⇒ α + ß + γ  = 2/3 ----(2)

sum of product of two roots = c/a = -1/3

⇒ αß + ßγ + γα = -1/3----(3)

⇒ αßγ = -1/3 ---(4)

Now, 4α + 4ß + 4γ = 4(α + ß + γ) = 4 x 2/3 = 8/3 (From 2)

4α .4ß + 4ß.4γ + 4γ. 4α = 16(αß + ßγ + γα)

 = 16 x -1/3 = -16/3 (From 3)

4α. 4ß. 4γ = 64αßγ = 64 x -1/3 = -64/3 (From (4))

\(\therefore\) Required equation whose roots are 4 times than given roots is x³ - (sum of roots)x² + (sum of products of two roots)x - (products of roots) = 0

⇒ x³ - (8/3)x² = (16/3) x - \((-64/3)=0\)

⇒ 3x³ - 8x² - 16x + 64 = 0 which is required equation.

(a + b + c)² - (a - b + c)²

 = (a² + b²+c² + 2ab + 2bc + 2ac) - (a² + b² + c² - 2ab - 2bc + 2ac)

 = 4ab + 4bc

= 4(a + c)b

Let f(x)=x² −3x−m(m+3) 

Above polynomial can be written as, 

f(x)=x² −(m+3)x + mx − m(m+3) 

=x(x−m−3) + 3(x−m−3) 

=(x−m−3)(x+m) 

To find the zeroes of f(x), put f(x)=0 

(x−m−3)(x+m)=0 

x−m−3=0 or x=−m

Required zeros are (m+3),−m

Given polynomial is 2x³ + x² − 6x − 3 

The two roots of polynomial are −√3​,√3​ 

Let the other root be a

Product of the roots will be - \(\frac{-3}{2}\)

⇒ (-√3)(√3)a = \(\frac{3}{2}\)

⇒a = -\(\frac{1}{2}\) 

So the other zero is -\(\frac{1}{2}\)

Correct option (B) 0   

Explanation:

 Cos 90° = 0 

(a) 9

\(\frac{d}{dt}\)(6a²) = 3.6 ⇒ 12a \(\frac{da}{dt}\) = 3.6

a\(\frac{da}{dt}\) = 0.3

\(\frac{dv}{dt}\) = \(\frac{d}{dt}\)(a³) 

= 3a \((a\frac{da}{dt})\)

= 3 x 10 x 0.3 

= 9

(4) 30

n(X_i) = 10. U_{i = 1} X_i for i ∈ 50 = T,

n (T) = 500 each element of T belongs to exactly 20 elements of 

X_i = 500/20 = 25 distinct elements 

So 5n/6 = 25 

n = 30