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Acidity of a base is the number of a mono basic acid required to neutralize 1 molecule of a base. [The base which contain number of replaceable – OH group].

Innate immunity is present at the time of birth. This is accomplished by providing different types of barriers to the entry of the foreign agents into our body.

It consists of four type of barriers:

1. Physical barrier - Skin and mucus

2. Physiological barrier - Acid in the stomach, saliva in the mouth and tears.

3. Cellular barrier - Neutrophils and macrophages.

4. Cytokine barrier - Interferons.

(i) Four because the zeros between the non-zero digits are significant figures. 

(ii) Four because only the first term gives the significant figures and exponential term is not considered. 

(iii) Four. However, if expressed in scientific notation as 8 × 10³, it will have only one significant figure, as 8.0 × 10³, 8.00 × 10³ or 8.000 x 10³, it will have 2, 3 or 4 significant figures. 

(iv) Two because the zeros on the left of the first non-zero digit are not significant. 

(v) As π = 22/7 = 3.1428571…., hence it has infinite number of significant figures. 

(vi) Three, because the reported sum will be only upto one decimal place i.e., 18.9. 

(vii) Two, because the number with least number of significant figures involved in the calculation (i.e., 14) has two significant figures.

Internal indicator are the substance which are added in the titration flask at the beginning. Example: Methyl orange, starch, phenolphthalein, diphenylamine. 

Internal indicator are further divided according to the type of reaction taking place during titration. 

(a) Acid base indicator: The choice of indicator for acid base titration depends on the nature of the acid and base used in that reaction. 

Acid Base Indicator 

Strong acid Strong base Methyl orange or phenolphthalein

(b) Redox indicator: Example: Diphenylamine. When titrating KMnO₄ with oxalic acid.

(d) 6.4 × 10^-5

Given, 0.000064 = 0. 64 x 10^-4 =6.4 x 10^-5

Hence, standard form of 0.000064 is 6.4 x 10^-5.

Molar mass : Molar mass of a substance is the mass of one mole of that substance.

Formula mass : Formula mass of a substance is the sum of the atomic masses of all atoms present in one formula unit of the compound (normally ionic compounds)

Correct answer is 1/50 N

(a) 2.34 × 10⁸

Explanation: 

234000000 

= 234 × 10⁶ 

= 2.34 × 10² × 10⁶ 

= 2.34 × 10⁸

(c) 1 Since, a⁰ = 1 (by law of exponent)

Correct option is (D) 2/5

\((\frac{2}{5})^{-3}\times(\frac{25}{4})^{-2}\) \(=(\frac{2}{5})^{-3}\times(\frac{5^2}{2^2})^{-2}\)

\(=((\frac52)^{-1})^{-3}\times((\frac{5}{2})^2)^{-2}\)

\(=(\frac52)^{-1\times-3}\times(\frac{5}{2})^{2\times-2}\)

\(=(\frac52)^{3}\times(\frac{5}{2})^{-4}\)

\(=(\frac52)^{3-4}\)

\(=(\frac52)^{-1}\)

= \(\frac{2}{5}\)

 Correct option is   D) \(\frac{2}{5}\)

Correct answer is 80%

Correct option is (B) x^-2 = √x

(A) \((x^{-3})^2=(x)^{-3\times2}=x^{-6}\) (True)

(B) \(x^{-2}=\frac1{x^2}\neq\sqrt{x}\) (Not true)

(C) \(\cfrac{x^{-3}}{x^{-2}}=x^{-3}.(x^{-2})^{-1}\) \(=x^{-3}x^2=x^{-3+2}=x^{-1}=\frac{1}{x}\) (True)

(D) \(x^{-3}\times x^{-5}=x^{-3+(-5)}=x^{-3-5}=x^{-8}\) (True)

Correct option is  B) x^-2 = √x 

Correct option is (D) 1

\(a^m\times a^{-m}=a^{m-m}=a^0=1\)

Correct option is  D) 1

Correct option is (C) 438 x 10⁵

43800000 = \(438\times10^5\)

C) 43.8 × 10⁵

Correct option is (B) \(\frac{1}{a^n}\)

\(a^{-n}\) = \(\frac{1}{a^n}\)

Correct option is   B) \(\frac{1}{a^n}\)

Correct option is (A) 3¹¹

\(3^5\div3^{-6}\) \(=3^5\times(3^{-6})^{-1}=3^5\times3^6\) \(=3^{5+6}\) \(=3^{11}\)

Correct option is  A) 3¹¹ 

Correct option is (A) \(\cfrac{a^m}{b^m}\)

\((\frac{a}{b})^m\) = \(\cfrac{a^m}{b^m}\)

Correct option is   A) \(\frac{a^m}{b^m}\)

Correct option is (C) 2/5

\(\left[(\frac{2}{5})^{-1}\right]^{-1}=(\frac{2}{5})^{-1\times-1}=(\frac{2}{5})^1\) \(=\frac{2}{5}\)

Correct option is  C) 2/5

Correct option is (D) 5

\((100^0+2^{-1}+1^{-2})\div2^{-1}\) \(=(1+\frac12+1)\times(2^{-1})^{-1}\)

\(=(2+\frac12)^2\)

\(=\frac{4+1}2\times2\)

\(=\frac{5}2\times2=5\)

Correct option is D) 5

Correct option is (C) 1

\(\cfrac{a^m}{a^m}=a^{m-m}=a^0=1\)

Correct option is  C) 1

Correct option is (C) 4

\(2^{-4}=\frac{1}{2^n}=2^{-n}\)

\(\Rightarrow\) -n = -4

\(\Rightarrow\) n = 4

Correct option is  C) 4

Correct option is (A) 9/16

\((\frac{4}{3})^{-2}=\left[(\frac{4}{3})^{-1}\right]^2=(\frac34)^2\) \(=\frac{3^2}{4^2}=\frac{9}{16}\)

Correct option is   A) \(\frac{9}{16}\)

Correct option is (A) 4

\((-2)^2=(-1)^2\times2^2=1\times4=4\)

Correct option is  A) 4

Correct option is (C) 437 x 10^-2

\(0.000437\times10^4\) = 4.37 \(=\frac{437}{100}\)

= \(437\times10^{-2}\)

C) 473 × 10^-2 

Correct option is (D) 1

\(100^0\) = 1 \((\because a^0=1,a\in\mathbb{R}-\{0\})\)

Correct option is  D) 1

Correct option is (A) –1

\((-9)^3\div9^3=\frac{(-9)^3}{9^3}=\frac{(-1)^3\times9^3}{9^3}\) \(=(-1)^3=-1\)

Correct option is  A) – 1

Correct option is (A) na

a + a + … + a   (n times) = na

Correct option is  A) na

Correct option is (A) \(\frac{1}{a^{n-m}}\)

\(\because\) m < n

\(\Rightarrow\) n - m > 0

\(\frac{a^m}{a^n}=\frac1{a^n\times a^{-m}}=\frac1{a^{n-m}}\)

Correct option is   A) \(\frac{1}{a^{n-m}}\)

Correct option is (D) 3/2

\(2^{-1}+1^{-1}\) \(=\frac12+\frac11=\frac{1+2}2\)

= \(\frac{3}{2}\)

Correct option is  D) 3/2

Correct option is (B) –1

\(8^{x+1}\) = 1 = \(8^0\)

\(\Rightarrow\) x+1 = 0

\(\Rightarrow\) x = -1

Correct option is  B) – 1

Correct option is (B) a²⁰¹⁶

\(a\times a\times a\times....\times a\) (2016 times) \(=a^{2016}\)

Correct option is  B) a²⁰¹⁶

Correct option is (B) x^-5

\(x^7\div x^{12}=\frac{x^7}{x^{12}}\) \(=x^{7-12}=x^{-5}\)

Correct option is  B) x^-5 

Molecular mass is calculated by multiplying average atomic mass of each element by the number of its atoms and adding them together.

e.g.,

Molecular mass of carbon dioxide (CO₂) is calculated as follows :

Molecular mass of CO₂ = (1 × average atomic mass of C) + (2 × average atomic mass of O) 

= (1 × 12.0 u) + (2 × 16.0 u) 

= 44.0 u

Correct option is (D) 1275

\(1.275\times10^3\) \(=\frac{1275}{1000}\times1000\) = 1275

Correct option is  D) 1275

Correct option is (D) 46700

\(4.67\times10^4\) \(=467\times10^2\)

= 46700

Correct option is  D) 46700

Correct option is (D) 1

\((\frac{1}{2016})^0=1\)  \((\because a^0=1\,for\,a\in\mathbb{R}-\{0\})\)

Correct option is  D) 1

Given : 

Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g.

To find : 

Atomic mass of hydrogen in u

Calculation : 

1.66056 × 10^-24g = 1 u

∴ 1.6736 × 10^-24 g = x

x = \(\frac{1.6736\times 10^{-24}g}{1.66056\times 10^{-24}g/u}\)

= 1.008u

∴ The atomic mass of hydrogen in u = 1.008 u

Correct option is (D) 0.0000203

\(2.03\times10^{-5}\) \(=\frac{2.03}{10^5}=\frac{2.03}{100000}\) = 0.0000203

D) 0.0000203

True

For standard form, 203000 

= 203 x 10 x 10 x 10 

= 203 x 10³

= 2.03 x 10² x 10³

= 2.03 x 10⁵

Gram-molar volume : The volume occupied by one gram-molecular mass of any gas at NTP (0° C or 273 K and one atm or 760 mm of Hg pressure). Its value is 22.4 litre. It is also known as molar volume

The usual form of 2.39461 × 10⁶ is 2394610.

Explanation:

2.39461 × 10⁶ 

= 2.39461 × 10 × 10 × 10 × 10 × 10 × 10

= 239461 × 10

= 2394610

The usual form of 3.41 × 10⁶ is 3410000.

Explanation: 

3.41 × 10⁶ 

= 3.41 × 10 × 10 × 10 × 10 × 10 × 10

= 341 × 10 × 10 × 10 × 10

= 3410000

The standard form of 12340000 is 1.234 × 10⁷

Explanation: 

12340000 

= 1234 × 10⁴ 

= 1.234 × 10 ³ × 10⁴ 

= 1.234 × 10⁷

Correct option is (A) 1

\((2^5\div2^6)\times2=(\frac{2^5}{2^6})\times2=\frac12\times2=1\)

Correct option is  A) 1

The standard form of (1/100000000) is 1.0 × 10^-8

Explanation: 

(1/100000000) 

= 1/1×10⁸ 

= 1.0 × 10^-8

We know that,

a⁰ = 1

[2^–1 + 3^–1 + 4^–1]⁰ = 1

[4^-1 +3^-1 + 6^-2]^-1

= (1/4+1/3+1/6²)^-1

= [(9+12+1)/36]^-1

= (22/36)^-1

= (36/22)