Correct Answer: B. 8 mins

Correct Answer: C) Na₂S

Explanation

The correct answer is Option C, Na2S. When Na2S reacts with dilute HCl, it produces hydrogen sulfide gas (H2S). The H2S gas turns dichromate paper green, which is a test for reducing agents. The yellow precipitate is likely to be sulfur, which is formed when H2S reacts with oxygen in the air. Therefore, the reaction can be represented as follows: Na2S + 2HCl ? H2S + 2NaCl.

It is important to note that Na2S is a reducing agent, which means that it can donate electrons to other substances. Additionally, dilute HCl is an acid that can donate hydrogen ions (H+) when dissolved in water. When Na2S reacts with dilute HCl, it produces H2S gas, which is a reducing agent that can donate electrons to dichromate ions (Cr2O7^2-) in the dichromate paper, turning it green.

In summary, the correct option for this Chemistry question is Option C, Na2S. When Na2S reacts with dilute HCl, it produces H2S gas, which turns dichromate paper green, and a yellow precipitate of sulfur is formed. Na2S is a reducing agent that can donate electrons to dichromate ions in the dichromate paper.

Correct Answer: B) 168cm³

Explanation

First, let's write down the balanced chemical equation for the reaction between calcium trioxocarbonate (IV) and hydrochloric acid (HCl):

\(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\)

Now, we can determine the limiting reactant in this reaction. We have 0.0075 moles of calcium trioxocarbonate (IV) and 0.015 moles of HCl. According to the balanced equation, one mole of calcium trioxocarbonate (IV) reacts with two moles of HCl. So, we will divide the moles of each reactant by their respective stoichiometric coefficients:

\(\frac{0.0075 \, \text{moles} \, CaCO_3}{1} = 0.0075\)

\(\frac{0.015 \, \text{moles} \, HCl}{2} = 0.0075\)

Since both values are the same, the reaction goes to completion, and there is no limiting reactant.

To find the volume of the gas evolved at S.T.P (Standard Temperature and Pressure), we need to determine the moles of carbon dioxide (CO₂) produced. According to the balanced equation, one mole of calcium trioxocarbonate (IV) produces one mole of CO₂. Therefore, 0.0075 moles of calcium trioxocarbonate (IV) will produce 0.0075 moles of CO₂.

At S.T.P, one mole of any gas occupies a volume of 22.4 L (or 22400 cm³). To find the volume of 0.0075 moles of CO₂ gas evolved, we can use the following equation:

\(\text{Volume of gas} = \text{moles of gas} \times \text{molar volume at S.T.P}\)

\(\text{Volume of gas} = 0.0075 \, \text{moles} \times 22400 \, cm^3\, \text{per mole} = 168 \, cm^3\)

Thus, the volume of the gas evolved at S.T.P is168 cm³ (Option B).

Correct Answer: A. the products have less heat content than reactants

Correct Answer: D. cations are smaller than the parent atom

Correct Answer: A) Electrovalent

Correct Answer: C. Bind rubber molecules together

Correct Answer: D. sulphur(iv)oxide

Explanation

In the majority of power plants, sulfur appears due to coal burning which is used to generate electricity.

Correct Answer: D. relative tendencies of oxidation

Correct Answer: B) 2-methylbut-1-ene