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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 1

Answer:

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.

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Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g)  NH3(g)(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.

Answers

Answer:

a) N2 (g) + H2 = 2 NH3

b) You have to state the mass of hydrogen

Which of the following statements best explains why the atomic radius increases from top to bottom of the periodic table? A. The electronegativity decreases from top to bottom of the periodic table, so the atoms near the bottom have an increased capacity for electrons. B. For the atoms lower in the periodic table, there is more difference in charge between the electrons and the protons, which allows electrons to orbit farther from nucleus. C. The ionization energy decreases from top to bottom of the periodic table, so the atoms near the bottom have an increased capacity for electrons. D. For the atoms lower in the periodic table, the valence electrons are in higher energy levels and farther from the nucleus.

Answers

Answer:

D. For the atoms lower in the periodic table, the valence electrons are in higher energy levels and farther from the nucleus.

Explanation:

Atomic radius increases down the group because down the group, there an increase in the number of principle energy levels occupied. Now, these higher principal energy levels are made up of orbitals that are larger than the orbitals from the lower energy levels in size.

Therefore, the effect of this is that the greater number of principal energy levels will outweigh the increase in nuclear charge since nuclear charge also increases down the group and this in turn makes the atomic radius to increase as we go down the group.

Answer: C. For the atoms lower in the periodic table, the balance electrons are in higher energy levels and farther from the nucleus.

Explanation: As the valance electrons orbit farther from the nucleus the energy level increases from the top to the bottom of the periodic table. So the atoms lower in the periodic table, the balance electrons are in higher energy levels and farther from the nucleus, which result in an increase in the atomic radius.

How can you remove sand from salt? Which physical property of sand was used in the process?

Answers

Answer:

You could collect the mixture and pour it in water, stir it , ad filter out the sand. This uses the physical property of solubility.

Explanation:

The salt dissolved, the sand didn't.

Draw the Lewis structure for the polyatomic hydronium H3O cation. Be sure to include all resonance structures that

Answers

Answer:

Lewis structure of Hydronium ion is shown below :

Explanation:

Lewis structure : It is a representation of valence electrons on the atoms in a molecule

Here , Hydronium ion is given , which contains 1 atom of oxygen and 3 atoms of hydrogen .

Oxygen has a total of 6 valence electrons and hydrogen contains 1 valence electron .

Oxygen share its 3 valence electrons with 3 hydrogen atoms and left with 3 valence electrons. From these three valence electrons of oxygen atom two electrons will be shown as a pair of electrons on oxygen atom but a single electron can not be shown . So , to simplify this, one positive charge is shown overall .

Resonance structure will be same as the hybrid structure because all three atoms are same , that is hydrogen .

The value of the Solubility Product Constant for lead phosphate is ____________Write the reaction that corresponds to this Ksp value.

_______(Aq,S,L) +_______(Aq,S,L) <-------->_______(Aq,S,L) +_______(Aq,S,L)

Ksp values are found by clicking on the "Tables" link.

Use the pull-down menus to specify the state of each reactant or product.

If a box is not needed leave it blank.

Answers

Answer: The reaction for the $K_(sp)$ value of lead phosphate is given below and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio. It is expressed as $K_(sp)$

The chemical formula of lead phosphate is $Pb_3(PO_4)_2$

The equation for the hydration of the lead phosphate is given as:

$Pb_3(PO_4)_2(s)+H_2O(l)\rightarrow 3Pb^(2+)(aq.)+2PO_4^(3-)(aq.)$

The solubility product of lead phosphate is $3.0\rightarrow 10^(-44)$ . This means that it is highly insoluble in water as the solubility product is very very low.

Hence, the reaction for the $K_(sp)$ value of lead phosphate is given above and the value of solubility product for the same is $3.0\rightarrow 10^(-44)$

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answers

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

$m_(PbI_2)=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2) *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\n\nn_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\n\nn_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI) *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\n\nn_(KI)=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \n\nm_(PbI_2)=18.2gPbI_2$

Best regards.

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