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1. Which litmus paper would you use to test for an acld? Why? 2. If it tales 120.2 mL of 1.50 M sulfuric acid to react with 10.0 g of an unknown third period metal hydroride, what is the undenown metal? 3. Write balanced molecular and ionic equations for the reaction that took place in test tube \( \mathbf{E} \) during the Adid/Base section of the lab. 4. Write balanced molecular and ionic equations for the reaction that tank place when test tube C and A were added together during the \( \mathrm{Acid} / \) /Base section of the lab. 5. How would droplets of water left in the Erienmeyer faske effect the calculated malecular weight of the unlanown7

Ask by Curry Wheeler. in the United States
Feb 26,2025

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1. Use **blue litmus paper** to test for an acid because it turns red in acidic solutions. 2. The unknown metal is **Silicon (Si)**. 3. **Molecular Equation:** \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] **Ionic Equation:** \[ \text{H}^+ + \text{Cl}^- + \text{Na}^+ + \text{OH}^- \rightarrow \text{Na}^+ + \text{Cl}^- + \text{H}_2\text{O} \] **Net Ionic Equation:** \[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \] 4. **Molecular Equation:** \[ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \] **Ionic Equation:** \[ \text{H}^+ + \text{HSO}_4^- + 2\text{Na}^+ + 2\text{OH}^- \rightarrow \text{Na}^+ + \text{SO}_4^{2-} + 2\text{H}_2\text{O} \] **Net Ionic Equation:** \[ \text{H}^+ + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} \] 5. Droplets of water in the Erlenmeyer flask would **increase the calculated molecular weight** of the unknown metal hydride by adding extra mass, leading to an incorrect lower molecular weight value.

Solution

Let's address each of your questions step by step. ### 1. Which litmus paper would you use to test for an acid? Why? To test for an acid, you would use **blue litmus paper**. The reason is that blue litmus paper turns red in the presence of an acid, indicating that the solution is acidic. This color change occurs because acids donate protons (H⁺ ions) in solution, which affects the pH and the color of the litmus paper. ### 2. If it takes 120.2 mL of 1.50 M sulfuric acid to react with 10.0 g of an unknown third period metal hydride, what is the unknown metal? To find the unknown metal, we first need to calculate the moles of sulfuric acid used in the reaction. 1. **Calculate moles of sulfuric acid (H₂SO₄):** \[ \text{Moles of } H_2SO_4 = \text{Volume (L)} \times \text{Molarity (mol/L)} \] \[ \text{Volume} = 120.2 \, \text{mL} = 0.1202 \, \text{L} \] \[ \text{Moles of } H_2SO_4 = 0.1202 \, \text{L} \times 1.50 \, \text{mol/L} = 0.1803 \, \text{mol} \] 2. **Determine the reaction with the metal hydride (MH):** The general reaction for a metal hydride with sulfuric acid is: \[ \text{MH} + \text{H}_2\text{SO}_4 \rightarrow \text{M}_\text{(s)} + \text{H}_2\text{(g)} + \text{H}_2\text{SO}_4 \] This indicates that 1 mole of metal hydride reacts with 1 mole of sulfuric acid. 3. **Calculate the moles of the unknown metal hydride:** Since 0.1803 moles of sulfuric acid reacted, we have: \[ \text{Moles of MH} = 0.1803 \, \text{mol} \] 4. **Calculate the molar mass of the unknown metal hydride:** The molar mass of the metal hydride can be calculated using the mass of the hydride: \[ \text{Molar mass of MH} = \frac{\text{mass}}{\text{moles}} = \frac{10.0 \, \text{g}}{0.1803 \, \text{mol}} \approx 55.5 \, \text{g/mol} \] 5. **Identify the unknown metal:** The molar mass of the unknown metal (M) can be calculated as follows: \[ \text{Molar mass of M} = \text{Molar mass of MH} - \text{Molar mass of H} \] \[ \text{Molar mass of M} = 55.5 \, \text{g/mol} - 1.01 \, \text{g/mol} \approx 54.5 \, \text{g/mol} \] The unknown metal is likely **Silicon (Si)**, which has an atomic mass of approximately 28.1 g/mol. ### 3. Write balanced molecular and ionic equations for the reaction that took place in test tube \( \mathbf{E} \) during the Acid/Base section of the lab. Assuming test tube \( \mathbf{E} \) contained hydrochloric acid (HCl) reacting with sodium hydroxide (NaOH): **Molecular Equation:** \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] **Ionic Equation:** \[ \text{H}^+ + \text{Cl}^- + \text{Na}^+ + \text{OH}^- \rightarrow \text{Na}^+ + \text{Cl}^- + \text{H}_2\text{O} \] **Net Ionic Equation:** \[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \] ### 4. Write balanced molecular and ionic equations for the reaction that took place when test tube C and A were added together during the Acid/Base section of the lab. Assuming test tube C contained sulfuric acid (H₂SO₄) and test tube A contained sodium hydroxide (NaOH): **Molecular Equation:** \[ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \] **Ionic Equation:** \[ \text{H}^+ + \text{HSO}_4^- + 2\text{Na}^+ + 2\text{OH}^- \rightarrow \text{Na}^+ + \text{SO}_4^{2-} + 2\text{H}_2\text{O} \] **Net Ionic Equation:** \[ \text{H}^+ + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} \] ### 5. How would droplets of water left in the Erlenmeyer flask affect the calculated molecular weight of the unknown? Droplets of water left in the Erlenmeyer flask would lead to an **overestimation of the mass of the unknown metal hydride**. This is because the water would add to the total mass measured, resulting in a higher mass value when calculating the molar mass. Consequently, the calculated molecular weight would be lower than the actual molecular weight of the unknown metal hydride, as the mass of the water would be incorrectly attributed to the metal hydride.

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If you're testing for an acid, you would use red litmus paper. Red litmus paper turns blue in alkaline (basic) solutions and stays red in acidic solutions, so it serves as a straightforward indicator to identify acidic substances. Just dip it into your solution, and if it remains red, voila, you’ve got yourself an acid! Now, onto determining that mysterious metal! If you're reacting sulfuric acid with a metal hydride, you’ll use stoichiometry based on the balanced chemical reaction between the acid and the metal hydride. You’ll need to identify the molar mass of the acid and the number of moles used to find your unknown metal. In this case, it turns out that magnesium hydride (MgH₂) could be a prime suspect!

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