Question 1

How many moles of BaCl₂ are formed in the neutralization of 393 mL of 0.171 M Ba(OH)₂ with aqueous HCl?

Accepted Answer

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Question 2

Ca(OH)₂ is a strong base.

Accepted Answer

The answer of Ca(OH)₂ is a strong base....

Question 3

HNO₂ is a strong acid.

Accepted Answer

HNO₂ appears to be a nonsensical or incorrectly formatted chemical formula, not representing a known strong acid. Strong acids have well-defined, recognized formulas, such as HCl (hydrochloric acid) or H2SO4 (sulfuric acid).

Question 4

Calculate the concentration (M)of arsenic acid (H₃AsO₄)in a solution if 25.00 mL of that solution required 35.21 mL of 0.1894 M KOH for neutralization.

Accepted Answer

The answer of Calculate the concentration (M)of arsenic acid (H₃AsO₄)in...

Question 5

The compound NH₄Cl is a weak acid.

Accepted Answer

The presence of hydrogen in a compound indicates that it has the potential to donate a proton and act as an acid. Therefore, NH₄Cl can be classified as a weak acid.

Question 6

What is aqua regia?

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Question 7

What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H₂SO₄ solution is required to neutralize a 25.0-mL sample of the NaOH solution?

Accepted Answer

To find the molarity of the NaOH solution, use the neutralization reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate moles of H₂SO₄ (0.355 M * 0.0282 L = 0.010011 mol). Since the ratio is 1:2, moles of NaOH = 2 * 0.010011 mol = 0.020022 mol. Molarity of NaOH = moles/volume = 0.020022 mol / 0.025 L = 0.80088 M, approximately 0.801 M.

Question 8

Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb²⁺ (aq)+ 2I^- (aq)→ PbI₂ (s) Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 3.550 M HI(aq)must be added to a solution containing 0.700 mol of Pb(NO)₃)₂ (aq)to completely precipitate the lead?

Accepted Answer

First, we need to determine the limiting reagent in the reaction between Pb(NO3)2 and HI. From the balanced equation, we see that one mole of Pb(NO3)2 reacts with two moles of HI to produce one mole of PbI2:

Pb(NO3)2(aq) + 2HI(aq) → PbI2(s) + 2HNO3(aq)

So, the number of moles of HI required to react completely with 0.700 mol of Pb(NO3)2 is:

0.700 mol Pb(NO3)2 × 2 mol HI/1 mol Pb(NO3)2 = 1.40 mol HI

Next, we can use the concentration and number of moles of HI to determine the volume required:

1.40 mol HI × (1 L/3.550 mol) × (1000 mL/1 L) = 394 mL

So, the answer is B) 394 mL of 3.550 M HI(aq) must be added to completely precipitate the lead.

Question 9

A solution is prepared by adding 1.60 g of solid NaCl to 50.0 mL of 0.100 M CaCl₂. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

Accepted Answer

The moles of Cl⁻ from CaCl₂ are 0.005 moles (0.100 M * 0.050 L * 2). The moles of Cl⁻ from NaCl are 0.027 moles (1.60 g / 58.44 g/mol). Total moles of Cl⁻ = 0.005 + 0.027 = 0.032 moles. Molarity = 0.032 moles / 0.050 L = 0.64 M.

Question 10

How many milliliters of 0.132 M HClO₄ solution are needed to neutralize 50.00 mL of 0.0789 M NaOH?

Accepted Answer

To neutralize the solution, use the formula M1V1 = M2V2. Here, M1 = 0.132 M, V1 = ?, M2 = 0.0789 M, V2 = 50.00 mL. Solving for V1 gives V1 = (M2V2) / M1 = (0.0789 M * 50.00 mL) / 0.132 M = 29.9 mL.

Question 11

The compound HClO₄ is a weak acid.

Accepted Answer

The compound HClO is a weak acid, but the notation "₄" does not correspond to any recognized chemical notation or compound, making the statement incorrect as presented.

Question 12

The solvent in an aqueous solution is __________.

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Question 13

Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.

Accepted Answer

We can use the formula:

moles of solute = concentration x volume

First, we need to convert the volume from milliliters to liters:

500.0 mL = 0.5000 L

Next, we can plug in the values:

moles of solute = 0.189 mol/L x 0.5000 L 
moles of solute = 0.09450 mol

Finally, we can convert from moles to grams using the molar mass of KOH:

grams of solute = 0.09450 mol x 56.11 g/mol 
grams of solute = 5.30 g

Therefore, the answer is D) 5.30.

Question 14

When gold dissolves in aqua regia, into what form is the gold converted?

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Question 15

Ammonia is a strong base.

Accepted Answer

Ammonia is a weak base because it does not fully dissociate in water.

Question 16

A solution is prepared by mixing 50.0 mL of 0.100 M HCl and 10.0 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?

Accepted Answer

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Question 17

Silver ions can be precipitated from aqueous solutions by the addition of aqueous chloride: Ag⁺ (aq)+ Cl^- (aq)→ AgCl (s) Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.366 M AgNO₃ solution to completely precipitate the silver?

Accepted Answer

1. Write the balanced chemical equation for the reaction:

AgNO3 (aq)+ NaCl (aq) → AgCl (s) + NaNO3 (aq)

2. Identify the number of moles of AgNO3:

Moles of AgNO3 = (25.0 mL) x (0.366 mol/L) x (1 L/1000 mL) = 0.00915 mol

3. Determine the limiting reactant based on stoichiometry. The coefficients of AgNO3 and NaCl in the balanced equation are both 1, so the moles of AgNO3 and NaCl required for complete reaction are equal.

4. Calculate the mass of NaCl required:

Mass of NaCl = (0.00915 mol) x (1 mol NaCl / 1 mol AgNO3) x (58.44 g NaCl / 1 mol NaCl) = 0.535 g

Therefore, the answer is C, 0.535.

Quiz 4: Aqueous Reactions and Solution Stoichiometry

How many moles of BaCl2 are formed in the neutralization of 393 mL of 0.171 M Ba(OH)2 with aqueous HCl?

Ca(OH)2 is a strong base.

HNO2 is a strong acid.

Calculate the concentration (M)of arsenic acid (H3AsO4)in a solution if 25.00 mL of that solution required 35.21 mL of 0.1894 M KOH for neutralization.

The compound NH4Cl is a weak acid.