Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behavior of the electron. Each function is characterized by 3 quantum numbers: n, l, and ml. If the value of n = 3 ... The quantum number l can have values from ? to ? . ... The total number of orbitals possible at the n = 3 energy level is ? . If the value of l = 3 ... The quantum number ml can have values from to ? . ... The total number of orbitals possible at the l = 3 sublevel is ?? .
Answer:
1) The quantum number l can have values from
2 to 0
2)The total number of orbitals possible at the n = 3 energy level is 3'2=9
3) If the value of l = 3 ... The quantum number ml can have values from 3 to -3
The quantum number l determines the shape of the orbital. In this case, if the value of n is 3, then the quantum number l can have values from 0 to (3-1), which is 2.
The total number of orbitals possible at the n = 3 energy level can be determined using the formula 2l + 1. So, for l = 0, there is 1 orbital. For l = 1, there are 3 orbitals. And for l = 2, there are 5 orbitals. Therefore, the total number of orbitals possible at the n = 3 energy level is 1 + 3 + 5 = 9.
On the other hand, the quantum number ml represents the magnetic quantum number. It specifies the orientation of the orbital in space. The value of ml ranges from -l to +l. So, if the value of l is 3, then the quantum number ml can have values from -3 to +3.
The total number of orbitals possible at the l = 3 sublevel can be determined using the formula 2ml + 1. So, for ml = -3, there is 1 orbital. For ml = -2, there is 3 orbitals. For ml = -1, there is 5 orbitals. For ml = 0, there is 7 orbitals. For ml = 1, there is 5 orbitals. For ml = 2, there is 3 orbitals. And for ml = 3, there is 1 orbital.
Therefore, the total number of orbitals possible at the l = 3 sublevel is 1 + 3 + 5 + 7 + 5 + 3 + 1 = 25.
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Come up with a definition for density
Density measures how tightly packed particles are.
If particles are tightly packed together, they will be more dense.
If they are loosely together, they will be less dense.
However, a common mistake is thinking that if something
is more dense it means that it's heavier.
However, that's not the case.
It has to do with how particles are packed in an object.
Oxide is a combination of oxygen and another element. So is Water an oxide?
Answer:
Yah, it's a neutral oxide
Explanation:
[tex]{ \bf{2H_{2(g)} +O_{2(g)} \: →2H _{2}O _{(l)} }}[/tex]
What is the formula for tin (II) chlorate decahydrate?
Answer:
The formula is:
Sn(ClO3)2
Tin(II) Chlorate is also called stannous chlorate and is a white-colored solid. Hydrates are the addition of water molecules. Tin (II) chlorate decahydrate is represented by Sn(ClO₃)₂ . 10 H₂O.
What are hydrates?Hydrates are the chemical compounds used to represent the water molecule (H₂O) in a compound. The water molecules are added to the compound formula as a crystalline structure.
The element tin is represented by the symbol Sn and chlorate is represented as (ClO₃)₂. The formula also has decahydrate which means it has ten molecules of water that can be represented by 10 H₂O.
The stock nomenclature is used to give the formula for the compound. The overall formula of the compound after adding the individual symbol will give, Sn(ClO₃)₂. 10 H₂O.
Therefore, Sn(ClO₃)₂ . 10 H₂O is a formula for tin (II) chlorate decahydrate.
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f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration need to be for a precipitate to occur?
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
HELP ASAP 15 POINTS
Why was Dalton's theory of the atom incorrect?
A. Dalton theorized that atoms were indivisible but they are actually made of smaller parts.
B. Dalton theorized that had negative charges spread throughout them but they are actually in electron shells.
C. Dalton' theory was correct.
D. Dalton theorized that atoms were too small to see but they are not.
Answer:
Answer is A.
Explanation:
The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. :)
Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the same, 10 B. the same, 14 C. a different, 6 D. a different, 10
Answer:
Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons
Explanation:
The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.
When these electrons change orbit or level they release energy in the form of light, which is known as a photon.
NEED ANSWER FAST 50 POINTS
A mixture of copper sulfate and water is heated, leaving a residue of copper sulfate in the container. Which method was used to separate the mixture?
A. chromatography
B. evaporation
C. filtration
D. distillation
The method used to separate copper sulfate and water mixture was evaporation.
Explanation:
Chromatography is a separation technique used in labs. In this technique, there are two phases, the mobile phase, and the stationary phase. The phase in which the mixture is dissolved is termed as mobile phase and the phase which serves as a carrier through the system like, sheet, capillary, etc. is termed as mobile phase.Evaporation is a process in which the action of heat is employed to separate dissolved solids from liquid. Due to heat liquid gets evaporated leaving the solid behind.Filtration is a process in which insoluble particles are separated from the liquid by allowing them to pass through a porous substance such as filter paper. Distillation is a process used in the separation of the mixture of liquids with different boiling points.So, from this, we can conclude that the method used to separate copper sulfate and water mixture was evaporation.
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Answer:
Just did this on my test the answer is evaportion.
Explanation:
1.Q= {n: 7 <n<31}, list the members of the set Q
Q={x:x[tex]\epsilon[/tex]n,7<n<31}
[tex]\\ \sf\longmapsto Q=\left\{8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30\right\}[/tex]
You can write it like this too
[tex]\\ \sf\longmapsto Q=\left\{8,9......30,31\right\}[/tex]
Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
For an ideal gas condition, what is the mass (g) of N2 if the pressure is 2.0 atm, the volume is 25 mL and the temperature is 290 Kelvin.
Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
At standard temperature and pressure conditions, the volume of an ideal gas contained in a jar is 55.3 L. How many molecules are in the jar. This question is to be answered in scientific notation.(eg. 1.5 e5)
Answer:
1.49e24
Explanation:
Standars temperature and pressure are 273.15K and 1atm, respectively.
Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:
PV = nRT
PV / RT = n
Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.
Replacing:
PV / RT = n
1atm*55.3L / 0.082atmL/molK*273.15K = n
2.47 moles = n
Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.
As we have 2.47 moles:
2.47 mol × (6.022x10²³ molecules / 1 mole) =
1.49x10²⁴ molecules that are in the jar
In scientific notation:
1.49e24Using only sodium carbonate, Na2CO3, sodium bicarbonate, NaHCO3, and distilled water determine how you could prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3. The total molarity of the ions should be 0.20 M. The Ka of the hydrogen carbonate ion, HCO3 - , is 4.7 x 10-11 .
Answer:
Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
Explanation:
The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:
[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:
[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex] (1)
Also, we know that:
[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex] (2)
From equation (2) we have:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex] (3)
By entering (3) into (1):
[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]
[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]
[tex] [NaHCO_{3}] = 0.103 M [/tex]
Hence, the [Na_{2}CO_{3}] is:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]
Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:
[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]
[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]
Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
I hope it helps you!
The respiration rate of a goldfish is measured. The goldfish is then placed in cold water and the respiration rate is measured again. What is the INDEPENDENT variable?
Answer:
Temperature of the water
Explanation:
In every study, there must be independent and dependent variables. An independent variable is the variable that is changed in order to obtain a response. In this case, the temperature of the water is being changed, the response in this experiment is the respiration rate of the goldfish.
Thus the respiration rate of the goldfish is the dependent variable because it is controlled by the temperature of the water and changes accordingly.
Summarily, the independent variable is the temperature of the water while the dependent variable is the respiration rate of the goldfish.
A student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) Br2(g) When she introduced 0.143 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.108 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
Answer:
1.84 × 10⁻³
Explanation:
Step 1: Write the balanced equation
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
Step 2: Calculate the initial concentration of NOBr
0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:
M = 0.143 mol / 1.00 L = 0.143 M
Step 3: Make an ICE chart
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
I 0.143 0 0
C -2x +2x +x
E 0.143-2x 2x x
Step 4: Find the value of x
The equilibrium concentration of NOBr(g) was 0.108 M. Then,
0.143-2x = 0.108
x = 0.0175
Step 5: Calculate the concentrations at equilibrium
[NOBr] = 0.108 M
[NO] = 2x = 0.0350 M
[Br₂] = x = 0.0175 M
Step 6: Calculate the equilibrium constant (Kc)
Kc = [0.0350]² × [0.0175] / [0.108]²
Kc = 1.84 × 10⁻³
Please help me fast! Pleaseee
I think 2 answer is write
balance the following reaction:
C₃H1₂ +
02
CO2 +
H20
Answer:
C₃H1₂ +
Explanation:
just took the exam
Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M AgNO 3? [K sp(Ag 2CrO 4) = 1.1 × 10 –12] What is the concentration of the silver ion remaining in solution?
Answer:
A precipitate will form.
[Ag⁺] = 2.8x10⁻⁵M
Explanation:
When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:
Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)
Ksp is defined as:
Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]
Where the concentrations [] are in equilibrium
Reaction quotient, Q, is defined as:
Q = [Ag⁺]² [CrO₄²⁻]
Where the concentrations [] are the actual concentrations
If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,
The actual concentrations are -Where 500mL is the total volume of the solution-:
[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M
[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M
And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴
As Q > Ksp; a precipitate will form
In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:
[Ag⁺] = 0.06M - 2X
[CrO₄²⁻] = 0.165M - X
Where X is defined as the reaction coordinate
Replacing in Ksp expression:
1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]
Solving for X:
X = 0.165M → False solution. Produce negative concentrations.
X = 0.0299986M
Replacing, equilibrium concentrations are:
[Ag⁺] = 0.06M - 2(0.0299986M)
[CrO₄²⁻] = 0.165M - 0.0299986M
[Ag⁺] = 2.8x10⁻⁵M[CrO₄²⁻] = 0.135M
A student is using a coffee-cup calorimeter to determine the enthalpy change of the endothermic reaction of two aqueous solutions. After both solutions are added to the cup, the student neglects to put the lid on the cup. This would cause the magnitude of the calculated ΔH° value to be: the answer is: too small, since the solution will absorb heat from the room. But why? Wouldn't depend on if the reaction releases or absorbs heat. Wouldn't it be too large because heat escapes the cup? I'm so confused
Answer:
Explanation:
In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .
The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .
Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.
Compounds A and B (both C10H14) show prominent peaks in their mass spectrum at m/z 134 and 119. Compound B also shows a less prominent peak at m/z 91. On vigorous oxidation with chromic acid, compound A is nonreactive while compound B yielded terephthalic acid.
Required:
From this information, deduce the structures of both compounds, and then draw the structure of B.
How many equivalent resonance structures can be drawn for the molecule of SO3 without having to violate the octet rule on the sulfur atom
Answer:
3
Explanation:
Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.
There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.
Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.
We can arrive at this answer because:
The octet rule is a rule that states that an atom must reach stability when it has eight electrons in the valence layer.This means that in bonds that cause the donation or sharing of electrons between atoms, each atom has eight electrons in the valence layer.In chemistry, resonance is a term that refers to structures created to represent the donation or sharing of electrons between the atoms of a molecule.These structures can be arranged in different ways, as long as they respect the octet rule.In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.
These structures are shown in the figure below.
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What is the electron configuration for the transition metal ion in each of the following compounds?
[Ni(H2O)6]Br2
[Cr(H2O)4(NO2)2]I
Answer:
1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled
2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled
1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.
The structures are shown in the image attached.
A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.
Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.
I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2) and tetrahydrofuran (image 3).
All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.
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A galvanic cell is powered by the following redox reaction:
2Zn2+(aq) + N2H4(aq) 4OH-zn2+ right arrow(aq) 2Zn(s) + N2(g) + 4H2O(I)
1. Write a balanced equation for the half-reaction that takes place at the cathode.
2. Write a balanced equation for the half-reaction that takes place at the anode.
3. Calculate the cell voltage under standard conditions.
Arrange the following in order of increasing boiling point: CH4, CH3CH3, CH3CH2Cl, CH3CH2OH. Rank from lowest to highest. To rank items as equivalent, overlap them.
Answer:
In order from lowest to highest:
Methane < Ethane < Chloroethene < Methanol
i.e: CH4 < CH3CH3 < CH3CH2OH < CH3CH2Cl
Explanation:
Compounds with stronger molecular fore have higher boiling points, thus making the molecules more difficult to pull apart. The presence of chains also increases the molecular dispersion. The dipole force of ethanol makes it have a very high boiling point.
I'm positive this explanation would suffice. Best of luck.
The order of increasing boiling points of the substances listed is; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH.
Intermolecular interactions occur between molecules. The boiling point and melting points of substances depends on the nature and magnitude of intermolecular interaction between the molecules of the substance.
The order of increasing boiling points of the substances listed is as follows; CH4 < CH3CH3 < CH3CH2Cl < CH3CH2OH. CH3CH2OH has the highest boiling point due to intermolecular hydrogen bonds in the molecule. Though CH4 and CH3CH3 are both alkanes, CH3CH3 has a higher molecular mass and consequently greater dispersion forces and a higher boiling point.
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Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.
Answer:
whats the ph ofpoh=9.78
Explanation:
what is the colour before and after when bromine reacts with chlorine ??
ANSWER
I need great answers
EXPLANATION
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Why is the r.a.m. Value used?
Answer:
The r.a.m. value is used to determine how many isotopes an element has.
Explanation:
a sample of oxygen occupied 568 ml. when the pressure increased to 911.6 mm Hg. at constant temperature, what volume did the gas initially occupy when the pressure was 248.71 mm Hg?
a 2082
b 155.0
c 2114
d 399.2
Answer:
[tex]\boxed {\boxed {\sf A. \ 2082 \ mL}}[/tex]
Explanation:
We are asked to find the initial volume of a gas given a change in pressure. Since the temperature remains constant, we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:
[tex]P_1V_1= P_2V_2[/tex]
The pressure was initially 248.71 millimeters of mercury, but the volume is unknown.
[tex]248.71 \ mm \ Hg * V_1 = P_2V_2[/tex]
The pressure is increased to 911.6 millimeters of mercury and the volume is 568 milliliters.
[tex]248.71 \ mm \ Hg * V_1 = 911.6 \ mm \ Hg * 568 \ mL[/tex]
We are solving for the initial volume, so we must isolate the variable V₁. It is being multiplied by 248.71 millimeters of mercury. The inverse operation of multiplication is division, so we divide both sides by 248.71 mm Hg.
[tex]\frac {248.71 \ mm \ Hg * V_1 } {248.71 \ mm \ Hg}= \frac{911.6 \ mm \ Hg * 568 \ mL }{248.71 \ mm \ Hg}[/tex]
[tex]V_1 = \frac{911.6 \ mm \ Hg * 568 \ mL }{248.71 \ mm \ Hg}[/tex]
The units of millimeters of mercury (mm Hg) cancel.
[tex]V_1 = \frac{911.6 * 568 \ mL }{248.71 }[/tex]
[tex]V_1 = \frac {517788.8} {248.71 } \ mL[/tex]
[tex]V_1 = 2081.897793 \ mL[/tex]
Let's round to the nearest whole number. The 8 in the tenths place tells us to round the 1 up to a 2.
[tex]V_1= 2082 \ mL[/tex]
The gas initially occupied a volume of 2082 milliliters and choice A is correct.
The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Answer:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Explanation:
Hello,
In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Best regards.