No

7.9 mg of Ca = 7.9\( \times\)10³ g of Ca
Atomic mass of Ca = 40.1
Moles of Al  =  \(\frac{ mass\ of\ Ca}{atomic\ mass}\)
= \( \frac{7.9\times10^{-3}}{40.1}\)
=  1.97\(\times \)10⁴ mol

For the reaction
\({ H }_{ 2 }{ SO }_{ 4 }+2NaOH{ \longrightarrow }{ Na }_{ 2 }{ SO }_{ 4 }+H_{ 2 }O\) 
1L: of 0.1 M H₂SO₄ contains = 0.1 mole of H₂SO₄
1L of 0.1 M Na OH contains = 0.1 mole of NaOH
According to the reaction,1 mole of  H₂SO₄ reacts with 2 moles of NaOH. Hence,0.1 mole of NaOH will react with 0.05 mole of H₂SO₄ 
(and 0.05 mole of H₂SO₄ will be left unreacted), i.e NaOH is the limiting reactant. Since, 2 moles of NaOH produces 1 mole of Na₂SO₄.
Hence,0.1 mole of NaOH will produces 0.05 mole of Na₂SO₄ 
Mass of Na₂SO₄ = moles \(\times \)molar mass
= 0.5\(\times \) (46 + 32  + 64)g = 7.10g
Volume of solution after mixing = 2L
Since only 0.05 mole of H₂SO₄ is left behind. as NaOH is completely used in the reaction.
Therefore,molarity of the given solution is calulated from moles of H₂SO₄.
H₂SO₄ left unreacted in the solution = 0.05 mole
\(\therefore \) Molarity of the solution \(=\frac { 0.05 }{ 2 } =0.025\)mol L^-1

Through the two values seem to be equivalent but scientifically they are different. 5.0 g has two significant figures and hence, its precision is 0.1 part in 5, i.e. 20 ppt 5.00 has three significant figures and hence, its precision is 0.01 parts in 5, i.e. 2 ppt. Hence, 5.00 g is more precise measurement than 5.0 g.

The  minimum uncertainty implies in this measurement is ± 0.01 cm.

On heating red oxide of copper in the presence of hydrogen, its mass decreases and it forms shining metallic copper which is chemically different from the original compound.
Cu₂O + H₂ ⟶ 2Cu + H₂O

It is because most of the elements occur in nature as a miture of isotopes and their atomic masses are the average relative atomic masses of the isotopes depending on their abundance.

The compound is glucose.Its molecular formula is C₆H₁₂O₆ while empirical formula is CH₂O.

Molecular mass of H₂O = 2 × 1 + 16 = 18 g mol^-1
18g of water = 1 gram of molecule 
∴ 576 g of water =1/18 × 576 = 32 gram molecule

\(1.3\times { 10 }^{ -22 }g\)

Fe = 28% ; s = 24%

14 g

0.33 mol

(a) 5
(b) 5
(c) 1
(d) 4

1.54 ×10³pm μ s ^-1.

Rounding off a number means that the digits which are not significant have to be dropped. The rules are as follows:
1. If the digit to be dropped is more than 5, then add 1 to the preceding significant figure. For e.g., in the number 11.06, the digit to be dropped is 6. Therefore, the preceding digit 0 is increased by one & the result is 11.1.
2. If the digit to be dropped is less than 5, then it is deleted as such without bringing any change in the preceding significant figure. For e.g., 43.123 is to report up to four significant figures, then the last digit 3 is dropped & the final answer is 43.12.
3. If the digit to be dropped is 5 then the preceding significant digit in the number may be left unchanged if it is even & may be increased by 1 in case it is odd. e.g., 1.6145, if reported up to four significant figures will become 1.614 after rounding off. Similarly, 1.6175 will become 1.618.
4. If during rounding off, more than one digit is to be dropped from a particular number, then they are dropped on at a time by following the above rules.
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810

(a) Calculation for molar of the gas
10.0 L of the given gas at STP weigh = 11.6 g
\(\therefore \) 22.4 L of the given gas at STP will weigh
\(\frac { 11.6\times 22.4 }{ 10 } =25.984\quad g\)
Molar mass = 25.984 = 26 \({ mol }^{ -1 }\)
(b) Empirical formula mass (CH) = 12 + 1 = 13
\(\therefore \) \(n=\frac { molecular \ mass }{ empirical \ formula \ mass } =\frac { 26 }{ 13 } =2\)
Hence, molecular formula 
\(=n\times CH=2\times CH={ C }_{ 2 }{ H }_{ 2 }\)

A = 55.41\(\mu\)

\(64\mu \) and \(32 \mu\)

mass of \({ O }_{ 2\\ }\)molecule = \(=5.32\times { 10 }^{ -23 }g\) ;
mass of \({ SO }_{ 2 }\) molecule \(=1.06\times { 10 }^{ -22 }g\)