Question 1

Why is sugar water not a good conductor of current?

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Question 2

Give an example of an inert electrode.

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Question 3

List the spectator ions in the following reaction: Ba⁺²(aq)+ 2OH^-(aq)+ 2 Na⁺(aq)+ CO₃^-2(aq)→ 2 Na⁺(aq)+ 2OH^-(aq)+ BaCO₃(s)

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Question 4

Balance the following equation: ________ C₉H₂₀ + ________ O₂→ ________ H₂O + ________ CO₂

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Question 5

Describe the difference between complete ionic and net ionic equations.

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Question 6

Balance the chemical equation given below,and determine the number of millilitres of 0.00300 mol L^-1 phosphoric acid required to neutralize 45.00 mL of 0.00150 mol L^-1 calcium hydroxide. ________ Ca(OH)₂(aq)+ ________ H₃PO₄(aq)→ ________ Ca₃(PO₄)₂(aq)+ ________ H₂O(l)

Accepted Answer

The balanced chemical equation is:

Ca(OH)2(aq) + 2H3PO4(aq) → Ca(H2PO4)2(aq) + 2H2O(l)

From the equation, we can see that one mole of Ca(OH)2 reacts with two moles of H3PO4.

First, we need to calculate the number of moles of Ca(OH)2:

Moles of Ca(OH)2 = volume x concentration
Moles of Ca(OH)2 = 0.045 L x 0.00150 mol/L
Moles of Ca(OH)2 = 6.75 x 10^-5 mol

Since two moles of H3PO4 are required to react with one mole of Ca(OH)2, we need:

Moles of H3PO4 = 2 x moles of Ca(OH)2
Moles of H3PO4 = 2 x 6.75 x 10^-5 mol
Moles of H3PO4 = 1.35 x 10^-4 mol

Finally, we can calculate the volume of 0.00300 mol/L H3PO4 needed:

Volume of H3PO4 = moles of H3PO4 / concentration
Volume of H3PO4 = 1.35 x 10^-4 mol / 0.00300 mol/L
Volume of H3PO4 = 0.045 L or 15.0 mL

Therefore, the answer is B) 15.0 mL.

Question 7

Balance the following equation: ________ C₁₀H₁₂ + ________ O₂→ ________ H₂O + ________ CO₂

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Question 8

If 100.mL of 0.400 mol L^-1 Na₂SO₄ is added to 200.mL of 0.600 mol L^-1 NaCl,what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive.

Accepted Answer

First, find the moles of Na⁺ and NaCl in their respective solutions: moles of Na⁺ = 0.400 mol/L x 0.100 L = 0.040 mol moles of NaCl = 0.600 mol/L x 0.200 L = 0.120 mol When the solutions are mixed, the moles of each ion will remain the same, but the total volume will increase. Thus: total moles of Na⁺ = 0.040 mol total moles of NaCl = 0.120 mol total volume = 0.100 L + 0.200 L = 0.300 L Therefore, the concentration of Na⁺ ions in the final solution is: 0.040 mol / 0.300 L = 0.133 mol/L However, Na^-1 is a divalent ion, meaning that each Na^-1 ion contributes two ⁺ ions. Thus, the concentration of ⁺ ions is: 0.133 mol/L x 2 = 0.267 mol/L However, the question asks for the concentration of Na⁺ ions, not ⁺ ions. Each Na⁺ ion also contributes one ⁺ ion. Since there are 0.040 mol of Na⁺ ions, there must be an additional 0.040 mol of ⁺ ions from the Na⁺ salt. Thus: total moles of ⁺ ions = 0.267 mol/L x 0.300 L + 0.040 mol = 0.110 mol concentration of ⁺ ions = 0.110 mol / 0.300 L = 0.367 mol/L Finally, since each NaCl molecule dissociates into one Na+ ion and one Cl- ion, the concentration of Na+ ions in the final solution is: 0.120 mol / 0.300 L = 0.400 mol/L Therefore, the concentration of Na⁺ ions in the final solution is: 0.367 mol/L + 0.400 mol/L = 0.767 mol/L Since this answer is not one of the choices, we must round to the appropriate number of significant figures: 0.767 mol/L rounded to three significant figures is 0.667 mol/L. Therefore, the best choice is B.

Question 9

How many millilitres of 0.550 mol L^-1 hydriodic acid are needed to react with 15.00 mL of 0.217 mol L^-1 CsOH? HI(aq)+ CsOH(aq)→ CsI(aq)+ H₂O(l)

Accepted Answer

First, we need to write and balance the chemical equation:

HI(aq) + CsOH(aq) → CsI(aq) + H2O(l)

From the balanced equation, we can see that the molar ratio between HI and CsOH is 1:1. This means that we need the same number of moles of HI as we have of CsOH to react completely.

To find the number of moles of CsOH, we use the formula:

moles = concentration (in mol/L) × volume (in L)

moles of CsOH = 0.217 mol/L × 0.01500 L = 0.003255 mol

Since the molar ratio between HI and CsOH is 1:1, we need the same number of moles of HI:

moles of HI = 0.003255 mol

Finally, we use the formula:

moles = concentration (in mol/L) × volume (in L)

moles of HI = 0.550 mol/L × volume of HI (in L)

volume of HI (in L) = moles of HI / concentration (in mol/L) = 0.003255 mol / 0.550 mol/L = 0.00592 L

To convert to milliliters:

volume of HI = 5.92 mL

Therefore, the answer is C: 5.92 mL.

Question 10

Based on the balanced chemical equation shown below,determine the mass percent of Fe³⁺ in a $0.7450 \mathrm {~g}$ sample of iron ore if 22.40 mL of a 0.1000 mol L^-1 stannous chloride,SnCl₂(aq),solution is required to completely react with theFe³⁺ present in the ore sample.The chemical equation for the reaction is 2Fe³⁺(aq)+ Sn²⁺(aq)? 2Fe²⁺(aq)+ Sn⁴⁺(aq)

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Question 11

Based on the balanced chemical equation shown below,determine the molarity of a solution containing Fe²⁺(aq)if 40.00 mL of the Fe²⁺(aq)solution is required to completely react with 30.00 mL of a 0.250 mol L^-1 potassium bromate,KBrO₃(aq),solution.The chemical equation for the reaction is 6Fe²⁺(aq)+ BrO₃^-(aq)+ 6H⁺(aq)→ 6Fe³⁺(aq)+ Br^-(aq)+ 3H₂O(l)

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Question 12

When 31.2 mL of 0.500 mol L^-1 AgNO₃ is added to 25.0 mL of 0.300 mol L^-1 NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq)+ NH₄Cl(aq)→ AgCl(s)+ NH₄NO₃(aq)

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Question 13

According to the balanced equation shown below,4.00 moles of oxalic acid,H₂C₂O₄,reacts with ________ moles of permanganate,MnO₄^-. 5H₂C₂O₄(aq)+ 2MnO₄^-(aq)+ 6H⁺(aq)→ 10CO₂(g)+ Mn²⁺(aq)+ 8H₂O(l)

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Question 14

How many grams of H₂ gas can be produced by the reaction of 54.0 grams of Al(s)with an excess of dilute hydrochloric acid in the reaction shown below? 2Al(s)+ 6HCl(aq)→ 2AlCl₃(aq)+ 3H₂(g)

Accepted Answer

Using stoichiometry, we can calculate the amount of H₂ gas produced. First, we need to find the limiting reactant. 54.0 g Al x (1 mol Al/26.98 g Al) = 2.00 mol Al Since the coefficient of Al is 2 and the coefficient of HCl is 6, we need 3 times as many moles of HCl to completely react with 2 moles of Al: 2.00 mol Al x (6 mol HCl/2 mol Al) = 6.00 mol HCl We have an excess of HCl, so it is not the limiting reactant. Next, we use the balanced equation to find the moles of H₂ produced: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) 2.00 mol Al x (3 mol H2/2 mol Al) = 3.00 mol H₂ Finally, we convert moles of H₂ to grams: 3.00 mol H₂ x (2.02 g H₂/mol) = 6.06 g H₂ Rounding to two significant figures gives an answer of 6.05 g, which is choice C.

Question 15

What is the molarity of a potassium triiodide solution,KI₃(aq),if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.200 mol L^-1 thiosulfate solution,K₂S₂O₃(aq)? The chemical equation for the reaction is 2S₂O₃^2-(aq)+ I₃^-(aq)→ S₄O₆^2-(aq)+ 3I^-(aq)

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Question 16

Identify oxidation and reduction half cell reactions: (a)Zn(s)→ Zn²⁺(aq)+ 2e⁻ (b)Cu²⁺(aq)+ 2e⁻ → Cu(s) Write the name of reaction.

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(a)Zn(s)→...

Question 17

Define a spectator ion.

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Question 18

Based on the balanced chemical equation shown below,what volume of 0.250 mol L^-1 K₂S₂O₃(aq)is needed to completely react with 24.88 mL of 0.125 mol L^-1 KI₃(aq),according to the following chemical equation: 2S₂O₃^2-(aq)+ I₃^-(aq)→ S₄O₆^2-(aq)+ 3I^-(aq).

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Question 19

What is the difference between a voltaic cell and an electrolytic cell?

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Question 20

How many millilitres of 0.200 mol L^-1 FeCl₃ are needed to react with an excess of Na₂S to produce 1.38 g of Fe₂S₃ if the percent yield for the reaction is 65.0%? 3Na₂S(aq)+ 2FeCl₃(aq)→ Fe₂S₃(s)+ 6NaCl(aq)

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Quiz 4: Chemical Reactions and Stoichiometry

Why is sugar water not a good conductor of current?

Give an example of an inert electrode.

List the spectator ions in the following reaction: Ba⁺²(aq)+ 2OH^-(aq)+ 2 Na⁺(aq)+ CO₃^-2(aq)→ 2 Na⁺(aq)+ 2OH^-(aq)+ BaCO₃(s)

Balance the following equation: C₉H₂₀ + O₂→ H₂O + CO₂

Describe the difference between complete ionic and net ionic equations.

Balance the chemical equation given below,and determine the number of millilitres of 0.00300 mol L^-1 phosphoric acid required to neutralize 45.00 mL of 0.00150 mol L^-1 calcium hydroxide. Ca(OH) ₂(aq) + H₃PO₄(aq) → Ca₃(PO₄) ₂(aq) + H₂O(l)

Balance the following equation: C₁₀H₁₂ + O₂→ H₂O + CO₂

If 100.mL of 0.400 mol L^-1 Na₂SO₄ is added to 200.mL of 0.600 mol L^-1 NaCl,what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive.

How many millilitres of 0.550 mol L^-1 hydriodic acid are needed to react with 15.00 mL of 0.217 mol L^-1 CsOH? HI(aq) + CsOH(aq) → CsI(aq) + H₂O(l)

When 31.2 mL of 0.500 mol L^-1 AgNO₃ is added to 25.0 mL of 0.300 mol L^-1 NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq) + NH₄Cl(aq) → AgCl(s) + NH₄NO₃(aq)

According to the balanced equation shown below,4.00 moles of oxalic acid,H₂C₂O₄,reacts with ________ moles of permanganate,MnO₄^-. 5H₂C₂O₄(aq) + 2MnO₄^-(aq) + 6H⁺(aq) → 10CO₂(g) + Mn²⁺(aq) + 8H₂O(l)

How many grams of H₂ gas can be produced by the reaction of 54.0 grams of Al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

Identify oxidation and reduction half cell reactions: (a)Zn(s)→ Zn²⁺(aq)+ 2e⁻ (b)Cu²⁺(aq)+ 2e⁻ → Cu(s) Write the name of reaction.

Define a spectator ion.

Based on the balanced chemical equation shown below,what volume of 0.250 mol L^-1 K₂S₂O₃(aq) is needed to completely react with 24.88 mL of 0.125 mol L^-1 KI₃(aq) ,according to the following chemical equation: 2S₂O₃^2-(aq) + I₃^-(aq) → S₄O₆^2-(aq) + 3I^-(aq) .

What is the difference between a voltaic cell and an electrolytic cell?

How many millilitres of 0.200 mol L^-1 FeCl₃ are needed to react with an excess of Na₂S to produce 1.38 g of Fe₂S₃ if the percent yield for the reaction is 65.0%? 3Na₂S(aq) + 2FeCl₃(aq) → Fe₂S₃(s) + 6NaCl(aq)

Quiz 4: Chemical Reactions and Stoichiometry

Why is sugar water not a good conductor of current?

Give an example of an inert electrode.

List the spectator ions in the following reaction: Ba+2(aq)+ 2OH-(aq)+ 2 Na+(aq)+ CO3-2(aq)→ 2 Na+(aq)+ 2OH-(aq)+ BaCO3(s)

Balance the following equation: ________ C9H20 + ________ O2 → ________ H2O + ________ CO2

Describe the difference between complete ionic and net ionic equations.

Balance the chemical equation given below,and determine the number of millilitres of 0.00300 mol L-1 phosphoric acid required to neutralize 45.00 mL of 0.00150 mol L-1 calcium hydroxide. ________ Ca(OH) 2(aq) + ________ H3PO4(aq) → ________ Ca3(PO4) 2(aq) + ________ H2O(l)

Balance the following equation: ________ C10H12 + ________ O2 → ________ H2O + ________ CO2

If 100.mL of 0.400 mol L-1 Na2SO4 is added to 200.mL of 0.600 mol L-1 NaCl,what is the concentration of Na+ ions in the final solution? Assume that the volumes are additive.

How many millilitres of 0.550 mol L-1 hydriodic acid are needed to react with 15.00 mL of 0.217 mol L-1 CsOH? HI(aq) + CsOH(aq) → CsI(aq) + H2O(l)

When 31.2 mL of 0.500 mol L-1 AgNO3 is added to 25.0 mL of 0.300 mol L-1 NH4Cl,how many grams of AgCl are formed? AgNO3(aq) + NH4Cl(aq) → AgCl(s) + NH4NO3(aq)

According to the balanced equation shown below,4.00 moles of oxalic acid,H2C2O4,reacts with ________ moles of permanganate,MnO4-. 5H2C2O4(aq) + 2MnO4-(aq) + 6H+(aq) → 10CO2(g) + Mn2+(aq) + 8H2O(l)

How many grams of H2 gas can be produced by the reaction of 54.0 grams of Al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

What is the molarity of a potassium triiodide solution,KI3(aq) ,if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.200 mol L-1 thiosulfate solution,K2S2O3(aq) ? The chemical equation for the reaction is 2S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3I-(aq)

Identify oxidation and reduction half cell reactions: (a)Zn(s)→ Zn2+(aq)+ 2e⁻ (b)Cu2+(aq)+ 2e⁻ → Cu(s) Write the name of reaction.

Define a spectator ion.

Based on the balanced chemical equation shown below,what volume of 0.250 mol L-1 K2S2O3(aq) is needed to completely react with 24.88 mL of 0.125 mol L-1 KI3(aq) ,according to the following chemical equation: 2S2O32-(aq) + I3-(aq) → S4O62-(aq) + 3I-(aq) .

What is the difference between a voltaic cell and an electrolytic cell?

How many millilitres of 0.200 mol L-1 FeCl3 are needed to react with an excess of Na2S to produce 1.38 g of Fe2S3 if the percent yield for the reaction is 65.0%? 3Na2S(aq) + 2FeCl3(aq) → Fe2S3(s) + 6NaCl(aq)

List the spectator ions in the following reaction: Ba⁺²(aq)+ 2OH^-(aq)+ 2 Na⁺(aq)+ CO₃^-2(aq)→ 2 Na⁺(aq)+ 2OH^-(aq)+ BaCO₃(s)

Balance the following equation: C₉H₂₀ + O₂→ H₂O + CO₂

Balance the chemical equation given below,and determine the number of millilitres of 0.00300 mol L^-1 phosphoric acid required to neutralize 45.00 mL of 0.00150 mol L^-1 calcium hydroxide. Ca(OH) ₂(aq) + H₃PO₄(aq) → Ca₃(PO₄) ₂(aq) + H₂O(l)

Balance the following equation: C₁₀H₁₂ + O₂→ H₂O + CO₂

If 100.mL of 0.400 mol L^-1 Na₂SO₄ is added to 200.mL of 0.600 mol L^-1 NaCl,what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive.

How many millilitres of 0.550 mol L^-1 hydriodic acid are needed to react with 15.00 mL of 0.217 mol L^-1 CsOH? HI(aq) + CsOH(aq) → CsI(aq) + H₂O(l)

When 31.2 mL of 0.500 mol L^-1 AgNO₃ is added to 25.0 mL of 0.300 mol L^-1 NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq) + NH₄Cl(aq) → AgCl(s) + NH₄NO₃(aq)

According to the balanced equation shown below,4.00 moles of oxalic acid,H₂C₂O₄,reacts with ________ moles of permanganate,MnO₄^-. 5H₂C₂O₄(aq) + 2MnO₄^-(aq) + 6H⁺(aq) → 10CO₂(g) + Mn²⁺(aq) + 8H₂O(l)

How many grams of H₂ gas can be produced by the reaction of 54.0 grams of Al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

Identify oxidation and reduction half cell reactions: (a)Zn(s)→ Zn²⁺(aq)+ 2e⁻ (b)Cu²⁺(aq)+ 2e⁻ → Cu(s) Write the name of reaction.

Based on the balanced chemical equation shown below,what volume of 0.250 mol L^-1 K₂S₂O₃(aq) is needed to completely react with 24.88 mL of 0.125 mol L^-1 KI₃(aq) ,according to the following chemical equation: 2S₂O₃^2-(aq) + I₃^-(aq) → S₄O₆^2-(aq) + 3I^-(aq) .

How many millilitres of 0.200 mol L^-1 FeCl₃ are needed to react with an excess of Na₂S to produce 1.38 g of Fe₂S₃ if the percent yield for the reaction is 65.0%? 3Na₂S(aq) + 2FeCl₃(aq) → Fe₂S₃(s) + 6NaCl(aq)