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What is evaporation? Calculate the solubility of sugar in water. Identify an acid salt among the given options. Determine the oxidation state of Mn in MnO2.

Chemistry

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Here are the solutions to the questions:

  1. The movement of liquid molecules from the surface of the liquid to the gaseous phase above it is known as evaporation. C. evaporation

  2. Step 1: Calculate the moles of sugar.

    Moles of sugar=massmolar mass=34.2 g342.0 g/mol=0.1 mol\text{Moles of sugar} = \frac{\text{mass}}{\text{molar mass}} = \frac{34.2 \text{ g}}{342.0 \text{ g/mol}} = 0.1 \text{ mol}

  1. Convert the volume of the solution from cm3\text{cm}^3 to dm3\text{dm}^3.

    $$
\text{Volume in dm}^3 = 50 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3} = 0.05 \text{ dm}^3
$$

  2. Calculate the solubility in mol/dm3\text{mol/dm}^3.

    $$
\text{Solubility} = \frac{\text{moles of sugar}}{\text{volume of solution}} = \frac{0.1 \text{ mol}}{0.05 \text{ dm}^3} = 2.0 \text{ mol/dm}^3
$$
D. $\boxed{\text{2.0 mol/dm}^3}$

    3. An acid salt is formed when a polyprotic acid is partially neutralized, meaning it still contains replaceable hydrogen atoms. * NaHSO4\text{NaHSO}_4 (sodium hydrogen sulfate) contains a replaceable hydrogen atom (HSO4−\text{HSO}_4^-), making it an acid salt. * Na2SO4\text{Na}_2\text{SO}_4, CH2CO2Na\text{CH}_2\text{CO}_2\text{Na} (sodium acetate), and Na2S\text{Na}_2\text{S} are normal salts as they do not contain replaceable hydrogen atoms. A. NaHSO4\text{NaHSO}_4

    1. Step 1: Determine the oxidation state of Mn in MnO2\text{MnO}_2. Let the oxidation state of Mn be xx. Oxygen has an oxidation state of -2. x+2(−2)=0x + 2(-2) = 0 x−4=0x - 4 = 0 x=+4x = +4

  3. Determine the oxidation state of Mn in MnCl2\text{MnCl}_2.

    Let the oxidation state of Mn be $y$. Chlorine has an oxidation state of -1.
$$
y + 2(-1) = 0
$$
$$
y - 2 = 0
$$
$$
y = +2
$$

  4. Calculate the number of electrons transferred.

    The oxidation state of Mn changes from +4 to +2. This is a decrease of 2, meaning 2 electrons are gained (reduction).  
A. $\boxed{2}$

    5. According to Le Chatelier's Principle, increasing pressure on an equilibrium system will shift the equilibrium towards the side with fewer moles of gas. Consider the reaction: ZnO(g)+H2(g)⇌Zn(s)+H2O(l)\text{ZnO(g)} + \text{H}_2\text{(g)} \rightleftharpoons \text{Zn(s)} + \text{H}_2\text{O(l)} * Reactant side: 1 mole of ZnO(g)\text{ZnO(g)} + 1 mole of H2(g)\text{H}_2\text{(g)} = 2 moles of gas. * Product side: Zn(s)\text{Zn(s)} (solid) + H2O(l)\text{H}_2\text{O(l)} (liquid) = 0 moles of gas. Since the product side has fewer moles of gas, increasing the pressure will shift the equilibrium to the right, favoring the formation of products. B. The equilibrium is driven to the right.

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