**(i) \(\dfrac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }\)**

Least precise no. of calculation = 0.112

Therefore, no. of significant numbers in the answer

= No. of significant numbers in the least precise no.

= 3

**(ii)** 5 × 5.365

Least precise no. of calculation = 5.365

Therefore, no. of significant numbers in the answer

= No. of significant numbers in 5.365

= 4

**(iii)** 0.012 + 0.7864 + 0.0215

As the least no. of decimal place in each term is 4, the no. of significant numbers in the answer is also 4.

Answered by Pragya Singh | 1 year agoChlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction:

4 HCl (aq) + MnO_{2}(s) → 2H_{2}O (l) + MnCl_{2}(aq) + Cl_{2} (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction, CaCO_{3} (s) + 2 HCl (aq) → CaCl_{2}(aq) + CO_{2} (g) + H_{2}O(l). What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl?

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Find:

**(i)** Empirical formula

**(ii)** Molar mass of the gas, and

**(iii)** Molecular formula

Calculate the number of atoms in each of the following

**(i) **52 moles of Ar

**(ii)** 52 u of He

**(iii)** 52 g of He

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope | Molar mass | Abundance |

\(\, _{ 36 }\textrm{Ar}\) | \( 35.96755 g \; mol^{ -1 }\) | 0.337 % |

\(\, _{ 38 }\textrm{Ar}\) | \( 37.96272 g \; mol^{ -1 }\) | 0.063 % |

\(\, _{ 40 }\textrm{Ar}\) | \( 39.9624g \; mol^{ -1 }\) | 99.600 % |