Suppose of lead(II) acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it. Round your answer to significant digits.

Answers

Answer 1
Answer:

Answer:

0.294 M

Explanation:

The computation of the final molarity of acetate anion is shown below:-

Lead acetate = Pb(OAc)2

Lead acetate involves two acetate ion.

14.3 gm lead acetate = Mass ÷ Molar mass

= 14.3 g ÷ 325.29 g/mol

= 0.044 mole

Volume of solution = 300 ml.

then

Molarity of lead is

= 0.044 × 1,000 ÷ 300

= 0.147 M

Therefore the molarity of acetate anion is

= 2 × 0.147

= 0.294 M

Answer 2
Answer:

Final answer:

To calculate the final molarity of acetate anion in the solution, consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. When ammonium sulfate is added, it reacts with the lead(II) cations, leaving only the acetate anions in the solution. The final concentration of acetate anions is therefore the same as the initial concentration.

Explanation:

To calculate the final molarity of acetate anion in the solution, we need to consider the dissociation of lead(II) acetate and the presence of ammonium sulfate. Lead(II) acetate will dissociate into lead(II) cations (Pb2+) and acetate anions (CH3COO-) in solution. However, when ammonium sulfate is added, the sulfate anions (SO42-) react with the lead(II) cations, forming lead(II) sulfate and removing them from solution. This leaves us with only the acetate anions.

First, calculate the concentration of the acetate anions in the lead(II) acetate solution. Then subtract the concentration of the acetate anions that reacted with the lead(II) cations to form lead(II) sulfate. This will give us the final concentration of acetate anions in the solution.

Let's assume we have an initial concentration of lead(II) acetate of X M. The dissociation of lead(II) acetate can be represented as:

Pb(CH3COO)2(s) ⇌ Pb2+(aq) + 2CH3COO-(aq)

Since we assume the volume of the solution doesn't change when the lead(II) acetate is dissolved, the initial concentration of acetate anions is also X M.

When ammonium sulfate is added, it reacts with the lead(II) cations according to the reaction:

Pb2+(aq) + SO4^2-(aq) ⇌ PbSO4(s)

Since the concentration of lead(II) sulfate is negligible, we can assume that all the lead(II) cations react with the sulfate anions. This removes the lead(II) cations from solution, leaving us with only the acetate anions.

Therefore, the final concentration of acetate anions is still X M.

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Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or negativea) Pb^2+(aq) + 2Cl-(aq) ---> PbCl2(s)b) CaCO3(s) ---> CaO(s) + CO2 (g)c) 2NH3(g) ---> N2(g) + 3H2(g)d) P4(g) + 5O2(g) ---> P4O10(s)e) C4H8(g) + 6O2(g) ---> 4CO2(g) + 4H2O(g)f) I2(s) ---> I2(g)

Answers

Answer: a) Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s):  negative

b)  CaCO_3(s)\rightarrow CaO(s)+CO_2(g) : positive

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g): positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s) : negative

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g):  positive.

f) I_2(s)\rightarrow I_2(g) : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) Pb^(2+)(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

As ions are moving to solid form , randomness decreases and thus sign of \Delta S is negative.

b)CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

c)2NH_3(g)\rightarrow N_2(g)+3H_2(g)

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of\Delta S is positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_(10)(s)

As gas is changing to solid, randomness decreases and thus sign of \Delta S is negative.

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of\Delta S is positive.

f)I_2(s)\rightarrow I_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

An aqueous solution is 4.44 M nitric acid and the density of the solution is 1.42 g/mL. Calculate the mole fraction of this solution.

Answers

The mole fraction of HNO3 is  0.225

Explanation:

1.Given data

Density = 1.429 /ml

Mass% = 63.01 g HNO3 / 100g of solution

The mass of 63.01 g is in 100 / 1.142 /ml of solution

Or 63.01 g in 55.7 mL

Molarity = 15.39 moles / L

Mass of water in 100g = 100 - 63.01=36.99 g

So 63.01 grams in 36.99 grams of water

So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703

Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles

Molality = 27.03 molal (mole / Kg)

Mole fraction = Mole of HN03 / Moles of water + mole of HNO3

Mole of water = 62/ 18 = 3.44

Moles of HNO3 = 63.01 / 63.01 = 1.000

Mole fraction = 1.000 / 3.44 + 1.000 = 0.225

The mole fraction of HNO3 is  0.225

Describe how you would make 250 ml of a 3 M solution of sodium acetate (NaOAc = 82.03 g/mol). First figure out how much sodium acetate you would need, then describe how you would make the solution if you were given a bottle of solid sodium acetate, a volumetric flask, and DI water.

Answers

Answer:

You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate

3 M (mol/L)*(1L/1000 mL)*(250 mL)*(82.03 g/1 mol)=61.52 g

Explanation:

First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.

What is my percent yield of titanium (II) oxide if I react 20 grams of titanium (II) oxide in excess water (that means TiS is my limiting reactant) and my actual yield of titanium (II) oxide is 13 g?

Answers

Answer:

your percent yield Is

Explanation:

1.5384615385

Brainliest plz

In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.

Answers

Answer:

This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.

Explanation:

In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.

Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.

So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.

Please Please! help help! so stress

Answers

I AM STRESSED AS WELL