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Bronsted-Lowry Acid

A Bronsted-Lowry acid, like an Arrhenius acid, is a compound that breaks down to give an H+ in solution.

Bronsted-Lowry Base

The Bronsted-Lowry base refers to any atom or ion capable of accepting or bonding to a free proton in solution.

examples include:

• Methanol – CH3OH
• Formaldehyde – H2CO
• And water – H2O

Lewis Acid

A Lewis acid refers to an atom or molecule that accepts an electron pair. 

Lewis Acid Examples in Organic Chemistry

• Borane – BH3 (hydroboration reaction)
• Aluminum Chloride – AlCl3 (electrophilic aromatic substitution reaction)
• Iron (III) Bromide – FeCl3 (electrophilic aromatic substitution reaction)
• and H+

Lewis Base

Since the Lewis definition has to do with the transfer of electrons, a Lewis Base is an electron pair donor. 

• A Lewis acid is an electron pair acceptor.

• A Lewis base is an electron pair donor.

The correct answer is : (a) 100

 (c) \(\vec {c}\) parallel to \(\vec {a}\)

{2,4,4,8,8,2} = {2,4,8}

Because if set elements are repeated then if written one time in roster from of set.

so given sets are equal.

f(x) = \(\sqrt{x-2}\) + \(\sqrt{x+47-14\sqrt{x-2}}\) ; x \(\geq\) 2

∴ f(50) = \(\sqrt{48}\) + \(\sqrt{50+47-14\times4\sqrt3}\)

 = 4\(\sqrt3+\sqrt{97-56\sqrt3}\)

f(66) = \(\sqrt{64}+\sqrt{66+47-14\sqrt{64}}\)

 = 8 + \(\sqrt{113-112}\)

 = 8 + 1

 = 9

∴ f(50) + f(66) = 4\(\sqrt3+\sqrt{97-56\sqrt3}+9\)

Since the happening of one of the given events A, B, C prevents the happening of other two, hence the given events are mutually exclusive.

Otherwise A ∩ B = ϕ, B ∩ C = ϕ, C ∩ A = ϕ

Hence they are mutually exclusive events.

Let y = 9^x

The mirror function of y = 9x about line y = x will find by inter changing x & y. i.e., putting x in place of y & y in place of x.

So, x = 9^y

⇒ y = log₉x   (By taking log₉ on both sides)

Hence, functions y = 9^x & y = log₉x are mirror images of each other about line y = x.

To find A × B we take one element from set A and one from set B. 

Given that 99 elements are common to both set A and set B.

Suppose these common elements are N₁​, N₂​, N₃​,... N₉₉​.

Select an ordered pair for A×B such that both are selected out of these common elements.

Examples: (N₁​,N₂​),(N₃​,N₅​)

All these will also be elements of B × A. 

Hence number of elements common to A × B and B × A is 99 × 99 = 99² ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)

n[(A×B)∩(B×A)] = n[(A∩B)∩(B∩A)] = (99)(99) = 99²

When a, b ∈ R and a ≥ 0, b ≥ 0 then √a > √b

⇔ a > b ≥ 0

∴ √(x + 2) > √(8 - x²)

⇔ x + 2 > 8 – x² ≥ 0 and x > –2, |x| < 2√2

We have (x + 2) > 8 – x²

⇔ x² + x – 6 > 0

⇔ (x + 3) (x – 2) > 0

⇔ x ∈ (– ∞, – 3) ∪ (2, ∞)

We have 8 – x² ≥ 0

⇔ x² ≤ 8 ⇔ |x| < 2√2

⇔ x ∈ [–2√2 , 2√2]

Also x + 2 > 0 ⇔ x > – 2

Hence x + 2 > 8 – x² ≥ 0

⇔ x ε ((– ∞, –3) ∪ (2, ∞)) ∩ [–2√2 , 2√2] and x > –2

⇔ x ∈ (2, 2√2)

The solution set is [x ∈ R : –2 < x ≤ 2√2]

y=(1+x)(1+x²)(1+x⁴)......(1+x²ⁿ)

⇒log y =Iog (1+x)+ Iog (1+x²) ......+ Iog (1+x²ⁿ)
⇒1/y. dy/dx= 1/(1+x )+ 2x/(1+x² )......+ 2nx²ⁿ⁻¹/(1+x²ⁿ)

⇒dy/dx= y.(1/1+x + 2x/1+x².......+ 2nx²ⁿ⁻¹/1+x²ⁿ)
Substitute the value x=0,
⇒∣dy/dx∣_x=0 =1.1 = 1

log y = log(1 + x) (1 + x²) (1 + x⁴).....(1 + x²ⁿ)

log y = log(1 + x) + log(1 + x²) + log(1 + x⁴) + ....+log(1 + x²ⁿ)

\(\frac1y\frac{dy}{dx} = \frac1{1+x}+\frac{2x}{1+x^2} + \frac{4x^3}{1+x^4}+....+\frac{2n x^{2n-1}}{1+x^{2n}}\) 

y at x = 0 is y(0) = 1

\(\therefore\) \(\frac{dy}{dx}\) at x = 0

\(\therefore\) (\(\frac{dy}{dx}\))_{(x = 0)} = y(0)(\(\frac1{1+0}+\frac{2\times0}{1+0}+\frac{4\times0}{1+0}\).....+\(\frac{2n-0}{1+0}\))

 = 1( 1 + 0 + 0 + .... + 0) = 1

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2. Used coconut shells can be used for craft to make musical instruments like the tabla, etc.
3. Pista shells can be used to fill in drawings adding beauty to the drawing.

In the given figure,

AC || DE because if we take DC as a transversal then angle ACD = angle CDE

But they are alternate angles,

Hence AC || DE

 ∵ ∠ EDC = 100° and ∠DCA are alternative interior angles of lines DE and AC whose transversal is CD.

∴ DE || AC (One pair of alternative angles are equal . so lines are parallel).

In any triangle, sum of angles is 180°.

\(\therefore \angle A + \angle B+ \angle C = 180° \)

⇒ \(\angle C = 180° - \angle A - \angle B\)

\(= 180° - 76° - 48° \)

\(= 180° - 124° \)

\(= 56 ° \)

The correct option is: (c)  πr²

Solution:

As the question is 16, 8, 12, 30, 105

also written as 2⁴, 2³ , 2²x3, 2x3x5, 3x5x7, 

As the power of 2 started ending and finally diminished and multiplication of prime number started.

So the next term must be 5x7x11 = 385

The value of cosec²θ -cot²θ = 1 

The common difference of A.P. 5, 9, 13, 17 is 1/3.

The value of sin 60°cos30° – sin30°cos60° is 1/2.

H.C.F. and L.C.M. of two numbers are 25 and 50 respectively. Which of the following is the product of the numbers 1250 

Which of the following is a rational number.√(49/64)  

The following is a solution of the quadratic equation 2x² + x – 6 = 0 x = 3/2

Centroid of a triangle divides its median in which of the following ratio 2 : 1 

(C) The value of c 

Correct option:

(B) 4,1 

Given f(x) = 2x – 5 

∴ (i) f(7) = 2(7) – 5 = 9. 

(ii) f(-3) = 2(-3) – 5 = -11.

Slope of the line = m 

= (y₂ - y₁)/(x₂ - x₁) 

= (4 - 2)/(-1 - 3) 

= 2/-4 

= -1/2

n(x ∪ y) = n(x) + (n ∪ y) – n(x ∩ y)

50 = 28 + 32 (x ∩ y) ⇒ [∴ n(X ∩ Y) = 10]

A – B = {1,3,5), (B – A) = {8}

• The sample space of four coins tossed together is 16.
• Each coin toss has 2 possible outcomes, i.e. H or T.
• So the flipping of 4 coins has 2x2x2x2 = 16 possible outcomes.

Sample space = {HH, HT,TT,TH}

~(√2 is not a complex no.) = √2 is a complex no.” 

Calculation:

Let the number be x

According to question,

8x = 40 – 8

⇒ 8x = 32

⇒ x = 4

Concept used:

Irrational numbers are those numbers which have decimal expansion that neither terminating nor repeating.

Calculation:

Like π = 3.1415926…….

√2 = 1.41421….

√3 = 1.73205……

But √4 = 2

Concept used:

For a number to be a perfect square, each prime factor has to be paired.

Calculation:

71232 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 53

The factors 3, 7 and 53 are not paired.

Hence, 71232 must be multiplied by 1113 i.e. (3 × 7 × 53) for it to be a perfect square

Given:

A number is divided by 3, the remainder will be 2

Calculation:

let the number be X

X = 3K + 2

According to question

X + 5 = 3K + 2 + 5

⇒X + 5 = 3K + 7

⇒X + 5 = 3K + 6 + 1

⇒X + 5 = 3(K + 2) + 1

Now,

3(K + 2) type number is definitely divisible by 3 so the remainder will be 1

Concept used:

HCF can be found by prime factorization method

HCF is the common factor of different numbers

Calculation:

88 = 2 × 2 × 2 × 11

220= 2 × 2 × 5 × 11

Common factor = 2 × 2 × 11

⇒ 44

∴ The HCF of 88 and 220 is 44

Let 

a = 0.232332333233332...

b = 0.212112111211112...

Here the decimal representation of a and b are non-terminating and non-repeating. So we observe that in first decimal place of a and b have the same digit 2 but digit in the second place of their decimal representation are distinct. And the number a has 3 and b has 1. So a > b.

Hence two rational numbers are 0.222, 0.221 lying between 0.232332333233332... and 0.212112111211112...

total 5 number of solutions i got this answer by drowning graph

Factor of 105 = 3*5*7

Factor of 154 = 2*7*11

HCF = 7