Answers
A 25.0 mL solution of Sr(OH)₂ is neutralized with 31.6 mL of 0.150 M HBr, then the concentration of the original Sr(OH)₂ solution is 0.189M.
How do we calculate the concentration?Concentration of the solution will be calculated by using the below chemical reaction as:
M₁V₁ = M₂V₂, where
M₁ & V₁ is the molarity and volume of HBr solution and M₂ & V₂ is the molarity and volume of original Sr(OH)₂ solution.
On putting values on above equation by taking from question, we get
M₂ = (0.15)(31.6) / (25) = 0.189 M
Hence required molarity of Sr(OH)₂ solution is 0.189M.
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Related Questions
What is the pH of a 0.0046 M nitric acid solution
Answers
Answer: 2.34
Explanation:
pH= -log (H+)
H+ is the acid's M
pH= -log (0.0046) =2.34
Question 4 (2 points)
Which best describes Nuclear changes?
The substance stays the same, but the properties change.
Elements rearranging to become different substances.
The number of protons or neutrons changes, which may result in a different
atom.
Answers
Nuclear modifications are alterations that take place inside an atom's nucleus. The amount of protons and neutrons in the nucleus may change, and this is the most fundamental degree of change that can take place in a material.
The atom is considered to have experienced a nuclear transition and is now a distinct atom when the number of protons or neutrons changes. This is thus because the element is determined by the number of protons, and the element changes if the number of protons varies.
If an atom of uranium contains 92 protons, for instance, it is uranium; nevertheless, if it has 91 protons, it is protactinium. This nuclear shift produces a distinct atom with different properties.
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Please help, its due today! I'll also make you brainiest (put them in an order that's simple, look at the picture and you'll see what I mean) Thank you and God bless! <33
On beaches there are often areas of grassy dunes where people are prohibited from walking. How do these protected areas preserve ecosystem services? Use the graphic organizer to categorize the following as either examples of land reclamation of protecting biodiversity.
Answers
Answer:
Preventing erosion – Land Reclamation
Protecting nesting areas – Protecting Biodiversity
Preventing littering – Land Reclamation
Preventing habitat disruption – Protecting Biodiversity
Protecting native species – Protecting Biodiversity
Preventing contamination of soil – Land Reclamation
Explanation:
I really hope I'm right! I tried my hardest, please give me brainliest :)
have a good day!
A 20.0 g sample of aluminum (specific heat = 0.902) g-1 oC-1) with an initial temperature of 2.5°C is heated with 427 J of
energy. What is the final temperature of the sample?
O 72.3°C
O 23.7°C
0 26.2°C
0 74.8°C
O 24.9°C
Answers
Answer:
T(final) = 26.2°C (3 sig. figs.)
Explanation:
Q = m·c·ΔT
m = mass of sample of interest = 20.0g
c = specific heat of sample of interest = 0.902j·g⁻¹·C⁻¹
ΔT = Temperature change = T(final) - T(initial) = T(f) - 2.5°C
Q = 427 joules
Q = m·c·ΔT => = ΔT = Q/m·c => T(f) = Q/m·c + T(i) = (427j/20.0g·0.902j·g⁻¹·C⁻¹) + 2.5°C = 26.16962306°C (calc. ans.) ≅ 26.2°C (3 sig. figs.)
17. Which of the following represents a formula for a chemical compound?A. CB. KOHC. O
Answers
KOH. Option B is correct
Explanations:A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.
The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)
Which atomic model is the one accepted by scientists today?
A) Dalton's model
B) Rutherford's model
C) quantum mechanical model
D) Bohr's model
Answers
what is stoichiometry?
Answers
Answer:
the relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.
Explanation:
G o o g l e
Explanation:
Stoichiometry is the calculation of reactants and products in chemical reactions in chemistry.
It takes a turtle 10 seconds to increase his speed from 1 m/s to 5 m/s. What is the acceleration of the turtle
Answers
The acceleration of the turtle, given the data from the question is 0.4 m/s²
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration v is the final velocity u is the initial velocity t is the time How to determine the accelerationThe acceleration of the turtle can be obtained as follow:
Time (t) = 10 sInitial velocity (u) = 1 m/sFinal velocity (v) = 5 m/sAcceleration (a) =?a = (v – u) / t
a = (5 – 1) / 10
a = 4 / 10
a = 0.4 m/s²
Thus, acceleration of the turtle is 0.4 m/s²
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How many HCl molecules and moles are there in 3.65 grams of
hydrogen chloride at normal conditions?
Answers
Answer:
0.6022×10²³ molecules
Number of moles = 0.1 mol
Explanation:
Given data:
Mass of HCl = 3.65 g
Number of molecules = ?
Number of moles = ?
Solution:
Number of moles:
Number of moles = mass/molar mass
Number of moles = 3.65 g/36.45 g/mol
Number of moles = 0.1 mol
Number of molecules:
1 mole of any substance contain 6.022×10²³ molecules
0.1 mol × 6.022×10²³ molecules/ 1 mol
0.6022×10²³ molecules
What is the PH for a solution that has an H+ ion concentration of 1.0x10^-6 M
Answers
Answer: POH=8
Explanation:PH = -log( H+concentration)
PH = 6
POH = 14 - PH = 8
Give two reasons why a luminous flame is not used for heating purposes. (2mks) a
Answers
Answer:
The Two reasons are :-
1. Combustion efficiency
2. Standardization of test conditions as non-luminous flame.
Explanation:
A luminous flame isn't suitable for heating as it gives out soot (A black powdery or flaky substance consisting largely of amorphous carbon, produced by the incomplete burning of organic matter)
Nitrogen-13 has a half-life of 20 minutes. how much of a 100 mg sample would remain after 60 minutes?
Answers
The amount of nitrogen-13 sample that remained after 60 minutes has been 25mg.
Half-life can be described as the time required by the substance to reduce half of its initial concentration.
The half-life of Nitrogen-13 has been 20 minutes. In 20 minutes, the sample will be reduced to half of its concentration,
The total time has been 60 minutes.
The number of half-life experienced by the sample has:
Number of half life= Total time/half life
Number of half life cycles= 60/20=3
The number of half-life cycles = 3
The sample has been reduced to 50% in the first half-life cycle and reduced to 25% by the end of 2nd half-life cycle.
The sample remained = 25% of the initial concentration.
The sample remained = 25/100.100 mg
The sample remained = 25 mg
Therefore, the amount of nitrogen-13 sample that remained after 60 minutes has been 25mg.
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A 500 mL gas sample is collected over water at a pressure of 740mmHg and 25°C. What is the volume of the dry gas at STP? (STP = 1 atm and 0°C) Vapor pressure at 25° of H2O equals 24mmHg.
Answers
1) List the known and unknown quantities.
Sample: gas.
Volume: 500 mL.
Pressure: 740 mmHg
Temperature: 25 ºC.
Vapor pressure at 25 ºC: 24 mmHg.
2) Pressure of the gas.
\(P_{gas}=P_{atm}-P_{water\text{ }vapor}\)\(P_{gas}=740\text{ }mmHg-24\text{ }mmHg\)\(P_{gas}=716\text{ }mmHg\)The pressure of the gas is 716 mmHg
3) Moles of gas
3.1- List the known quantities.
Volume: 500 mL.
Temperature: 25 ºC.
Pressure: 716 mmHg.
Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).
3.2- Set the equation.
\(PV=nRT\)3.3- Convert the units of the volume, the temperature, and the pressure.
Volume.
1 L = 1000 mL
\(L=500\text{ }mL*\frac{1\text{ }L}{1000\text{ }mL}=0.500\text{ }L\)Temperature.
\(K=25\text{ }ºC+273.15\text{ }K\)\(K=298.15\text{ }K\)Pressure
1 atm = 760 mmHg
\(atm=716\text{ }mmHg*\frac{1\text{ }atm}{760\text{ }mmHg}=0.942\text{ }atm\)3.4- Plug in the know quantities in the ideal gas equation.
\((0.942\text{ }atm)(0.500\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(298.15\text{ }K)\)3.5- Solve for n (moles).
Divide both sides by (0.082057 L * atm * K^(-1) * mol^(-1)) * (298.15 K)
\(\frac{(0.942atm)(0.500L)}{(0.082057\text{ }L*atm*K^{-1}mol^{-1})(298.15K)}=\frac{n(0.082057\text{ }L*atm*K^{-1}mol^{-1})(298.15K)}{(0.082057\text{ }L*atm*K^{-1}mol^{-1})(298.15K)}\)\(n=\frac{(0.942atm)(0.500L)}{(0.082057L*atm*K^{-1}*mol^{-1})(298.15K)}=\)\(n=0.0193\text{ }mol\)
4) Dry volume at STP
STP conditions are
Temperature: 273 K
Pressure: 1 atm.
At STP conditions 1 mol of a gas occuppies 22.4 L. We can use this as a conversion factor.
1 mol gas = 22.4 L
\(V=0.0193\text{ }mol\text{ }gas*\frac{22.4\text{ }L}{1\text{ }mol\text{ }gas}=0.432\text{ }L\)The volume of the dry gas at STP is 0.432 L.
.
Answer:
1) List the known and unknown quantities.
Sample: gas.
Volume: 500 mL.
Pressure: 740 mmHg
Temperature: 25 ºC.
Vapor pressure at 25 ºC: 24 mmHg.
2) Pressure of the gas.
The pressure of the gas is 716 mmHg
3) Moles of gas
3.1- List the known quantities.
Volume: 500 mL.
Temperature: 25 ºC.
Pressure: 716 mmHg.
Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).
3.2- Set the equation.
3.3- Convert the units of the volume, the temperature, and the pressure.
Volume.
1 L = 1000 mL
Temperature.
Pressure
1 atm = 760 mmHg
3.4- Plug in the know quantities in the ideal gas equation.
3.5- Solve for n (moles).
Divide both sides by (0.082057 L * atm * K^(-1) * mol^(-1)) * (298.15 K)
4) Dry volume at STP
STP conditions are
Temperature: 273 K
Pressure: 1 atm.
At STP conditions 1 mol of a gas occuppies 22.4 L. We can use this as a conversion factor.
1 mol gas = 22.4 L
The volume of the dry gas at STP is 0.432 L.
Explanation:
What is a neutralization reaction?
Answers
Plz help it over science stuff
Answers
Human reaction time is usually about 0.18 s. If your lab partner holds a ruler between your finger and thumb and releases it from rest without warning, how far can you expect the ruler to fall before you catch it? (a = -g = -9.81 m/s .) - 0.32 m ⊝ - 0.88 m ⊝ - 0.16 m ⊝ - 0.09 m ⊝ CLEAR ALL ❮ PREVIOUS
Answers
Answer:
–0.16 m
Explanation:
From the question given above, the following data were obtained:
Time (t) = 0.18 s
Acceleration due to gravity (g) = –9.81 m/s²
Height (h) =?
We can obtain how far the ruler will fall by using the following equation:
H = ½gt²
H = ½ × –9.81 × 0.18²
H = ½ × –9.81 × 0.0324
H = –0.16 m
Thus, the ruler will fall –0.16 m before you will catch it.
Write the solubility product expressions for the following compounds.
(a) Ag2CO3
(b) Hg2Cl2
Answers
(a) The solubility product expression for Ag2CO3 is:
Ksp = [Ag+]^2[CO3^2-]
(b) The solubility product expression for Hg2Cl2 is:
Ksp = [Hg2^2+][Cl^-]^2
list three possible experimental errors that may have caused deviation of your answer to 6 from the accepted value. for each error, indicate whether you expect the error would cause your calculated value of the mass to be larger or smaller than the correct value.
Answers
When you step on the scale, you might see a reading of 160 lbs, but in actuality, you might weigh 160.11111 pounds, that would be challenging to determine with a standard bathroom scale.
Describe scale.
A fish's external body covering, specifically, consists of a tiny, flattened, stiff, and unmistakably delimited plate. B: A tiny, thin plate with mica scales that resemble the scales of a moth's wing.
A little weighing scale – what is it?
It is a compact weighing scale with an easy-to-read digital screen that is portable. It is a compact weighing scale with an easy-to-read computer indicator that is portable. It is a versatile scale that works well for weighing big goods in various locations.
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What is the density of an unknown compound in g/ml if 1.28 pounds of the compound has a volume of 4.50L
Answers
1.28 pounds * 453.59 grams/pound = 580.61 grams
Next, we can use the formula for density:
Density = Mass / Volume
Density = 580.61 grams / 4.50 L
Density = 128.91 g/L
Therefore, the density of the unknown compound is 128.91 g/L or 0.12891 g/mL (since there are 1000 mL in 1 L).
What did J. J. Thomson's cathode ray experiment show about atoms?
Answers
Answer:
atoms contain tiny negatively charged subatomic particles or electrons
Can you please walk me through this question?A vinegar solution (acetic acid) has a hydroxide ion concentration of 0.000000000005 M at 25 degrees Celsius. What is the hydronium ion concentration, in molarity units, of the vinegar solution ?
Answers
The first step to solve this question is to find the pOH of the solution, to do it, we have to use the following formula:
\(pOH=-log[OH^-]\)Where [OH-] is the concentration of hydroxide ions.
\(\begin{gathered} pOH=-log0.000000000005 \\ pOH=11.3 \end{gathered}\)Now, we have to find the pH of the solution. Remember that the sum of pH and pOH is always 14, we can use this information to find the value of the pH:
\(pH+pOH=14\)\(\begin{gathered} pH=14-pOH \\ pH=14-11.3 \\ pH=2.7 \end{gathered}\)We can find the hydronium ion concentration using the following equation:
\(\begin{gathered} pH=-log\lbrack^H^+] \\ 10^{-\frac{}{}pH}=\lbrack H^+] \\ 10^{-2.3}=\lbrack H^+] \\ \lbrack H^+]=0.005M \end{gathered}\)It means that the concentration of hydronium ions is 0.005M
chemical equation for Mercury and oxygen react to yield mercuric oxide.
Answers
Answer:
mercury has oxide yes
Explanation:
The chemical equation for mercury and oxygen reacting to yield mercuric oxide is 2 Hg + O₂\(\rightarrow\) 2 HgO.
What is chemical equation?Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.
A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .
The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.
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Calculate the effect of adding H3O+ and OH- on a buffer solution consisting of A) 0.5M CH3COOH and 0.5M CH3COONa B)after adding 0.02 mol of solid NaOH to 1.0L of the buffer solution in part a Ka of CH3COOH= 1.8x10-5 assuming the addition caused negligible volume changes. C) after adding 0.02 mol of HCL to 1.0L of buffer solution in (A).
Answers
A) the addition of \(H_3O\)+ or OH- ions would have opposing effects on the buffer solution.
b) \(H_3O\)+ would favor the formation of acetic acid, while adding OH- would favor the formation of acetate ions.
c) The exact magnitude of the changes in concentrations depends on the initial concentrations of \(CH_3COOH\) and \(CH_3COONa\), as well as the specific amount of \(H_3O\)+ or OH- added.
To determine the effect of adding \(H_3O\)+ and OH- on the given buffer solution, we need to consider the ionization of acetic acid (\(CH_3COOH\)) and the dissociation of its sodium salt (CH3COONa). Let's analyze each scenario separately:
A) Buffer solution consisting of 0.5 M\(CH_3COOH\) and 0.5 M \(CH_3COONa\):
When acetic acid (CH3COOH) and its sodium salt (\(CH_3COONa\)) are present together in a solution, they form a buffer system. Acetic acid partially ionizes in water, releasing \(H_3O\)+ ions, while sodium acetate dissociates into Na+ and \(CH_3COO\)- ions.
Adding \(H_3O\)+:
The \(H_3O\)+ ions would react with the acetate ions (CH3COO-) to form undissociated acetic acid (\(CH_3COOH\)) through the following reaction:
\(H_3O\)+ + \(CH_3COO\)- ⇌ \(CH_3COOH\) + H2O
The addition of H3O+ would shift the equilibrium to the left, promoting the formation of more acetic acid and decreasing the concentration of acetate ions.
Adding OH-:
The OH- ions would react with the acetic acid (\(CH_3COOH\) to form water and acetate ions (CH3COO-) through the following reaction:
OH- + \(CH_3COOH\) ⇌ \(CH_3COO\)- + H2O
The addition of OH- would shift the equilibrium to the right, consuming acetic acid and increasing the concentration of acetate ions.
B) After adding 0.02 mol of NaOH to 1.0 L of the buffer solution:
When solid NaOH is added to the buffer solution, it dissociates completely in water to form Na+ and OH- ions.
NaOH dissociation:
NaOH → Na+ + OH-
The OH- ions formed would react with acetic acid according to the reaction mentioned in Scenario A (2), increasing the concentration of acetate ions and consuming acetic acid.
C) After adding 0.02 mol of HCl to 1.0 L of the buffer solution:
When HCl is added to the buffer solution, it dissociates completely in water to form \(H_3O\)+ and Cl- ions.
HCl dissociation:
HCl → \(H_3O\)+ + Cl-
The \(H_3O\)+ ions formed would react with acetate ions (\(CH_3COO\)-) according to the reaction mentioned in Scenario A (1), forming more undissociated acetic acid and decreasing the concentration of acetate ions.
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When the following solutions are mixed together, what precipitate (if any) will form?
a. Hg(NO,)(ag) + CuSO (ag)
b. Ni(NO,)(ag) + CaCh(ag)
c. K;COg(ag) + Mgl.(ag)
d. Na;CrO.(ag) + AlBs(ag)
Answers
The precipitates that will be formed respectively from the reactions would be:
\(HgSO_4\)No precipitate \(MgCO_3\) \(Al_2 (CrO_4)_3\)What are precipitation reactions?Precipitation reactions are reactions during which two aqueous salt solutions combine to produce one aqueous and one insoluble salt.
Following this definition, the precipitates that will be formed from each of the reactions can be deduced as follows:
\(Hg_2(NO_3)(aq) + CuSO_4 (aq) --- > HgSO_4 (s) + CuNO_3 (aq)\). The precipitate here is \(HgSO_4\)\(Ni(NO_3)_2(aq) + CaCl_2(aq) --- > NiCl_2 (aq) + Ca (NO_3)_2 (aq)\). No precipitate is formed here.\(K_2CO_3(aq) + Mgl_2(aq)--- > KI (aq) + MgCO_3 (s)\). The precipitate formed here is \(MgCO_3\)\(Na_2CrO_4(aq) + AlBr_3(aq) --- > NaBr (aq) + Al_2 (CrO_4)_3 (s)\). The precipitate formed here is \(Al_2 (CrO_4)_3\)The production of precipitates follows solubility rules. Some of the rules are:
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A. Identify the structure drawn below.
Answers
The given structure contains 6 carbon atoms and 14 hydrogen atoms. Therefore, it is the structure of hexane.
Hexane is a hydrocarbon compound with the chemical formula C₆H₁₄. It belongs to the alkane family, which means it consists of only single bonds between carbon atoms. It is a colorless and odorless liquid at room temperature and is commonly used as a solvent in various industries.
It is a straight chain of six carbon atoms (C) with attached hydrogen atoms (H). Each carbon atom in the chain is bonded to two hydrogen atoms except for the two end carbon atoms, which are bonded to three hydrogen atoms.
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Your question is incomplete, most probably the full question is this:
A. Identify the structure drawn below.
The image is attached below.
In which of the following options can competition occur?
b
с
Between two
organisms that are
attempting to use
different resources
that are in short
supply
Between two
organisms that are
attempting to use
the same resource
that is in short
supply
Between two
organisms that are
attempting to use
the same resource
that is in limitless
supply
Answers
Answer: no answer
Explanation:
no anwser
If 20.6 grams of ice at zero degrees Celsius completely change into liquid water at zero degrees Celsius, the enthalpy of phase change will be positive. TRUE FALSE
Answers
Answer:
TRUE.
Explanation:
Hello,
In this case, since the fusion enthalpy of ice is +333.9 J/g and the fusion entropy is defined as:
\(\Delta _{fus}S=\frac{m*\Delta _{fus}H}{T_{fus}}\)
We can compute it considering the temperature (0 °C) in kelvins:
\(\Delta _{fus}S=\frac{20.6g*333.9J/g}{(0+273)K}\\\\\Delta _{fus}S=25.2J/K\)
Therefore answer is TRUE.
Best regards.
Venus's atmosphere, while primarily CO2, is also 3.5% nitrogen gas (i.e. mole fraction of 0.035). What is the partial pressure of nitrogen on Venus in kPa given that the total atmospheric pressure is 1334 psi?
Answers
The partial pressure of nitrogen on Venus is approximately 321.914 kPa.
To find the partial pressure of nitrogen on Venus, we need to calculate the partial pressure using the mole fraction of nitrogen and the total atmospheric pressure. First, we convert the total atmospheric pressure from psi to kilopascals (kPa) since the mole fraction is given in terms of kPa.
1 psi = 6.89476 kPa
Therefore, the total atmospheric pressure on Venus is:
1334 psi × 6.89476 kPa/psi = 9197.53 kPa
Next, we can calculate the partial pressure of nitrogen using the mole fraction. The mole fraction of nitrogen is given as 0.035, which means that nitrogen makes up 3.5% of the total moles of gas in the atmosphere.
The partial pressure of nitrogen is given by:
Partial pressure of nitrogen = Mole fraction of nitrogen × Total atmospheric pressure
Partial pressure of nitrogen = 0.035 × 9197.53 kPa
Partial pressure of nitrogen = 321.914 kPa
Therefore, the partial pressure of nitrogen on Venus is approximately 321.914 kPa.
It's important to note that the given atmospheric composition of Venus's atmosphere and the total atmospheric pressure are approximate values and can vary depending on specific conditions and measurements.
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A high protein diet contains 70.0g of carbohydrates, 5.0g of Fat, and 150g of protein. How much energy, in kilocalories, and kilojoules, does each diet provide? Round answers off to tens place
Answers
Answer:
925kcal, 3870.2kJ.
Explanation:
The calorie densities of these energy sources are:
Carbohydrates: 4kcal/g
Fat: 9kcal/g
Protein: 4kcal/g
70.0g of carbohydrates are:70.0g * (4kcal/g) = 280kcal
5.0g fat * (9kcal/g) = 45kcal
150g protein * (4kcal/g) = 600kcal
The energy in kilocalories is 600+280+45 = 925kcal
As 1kcal is 4.184kJ. 925kcal are:
925kcal * (4.184kJ/1kcal) = 3870.2kJ
How many moles of aluminum ions al3+ are present in 0.42 mol of al2so43
Answers
There are 0.84 moles of aluminum ions (Al3+) present in 0.42 mol of Al2(SO4)3.
To determine the number of moles of aluminum ions (Al3+) present in 0.42 mol of Al2(SO4)3, we need to consider the stoichiometry of the compound.
The formula of aluminum sulfate (Al2(SO4)3) indicates that for every 1 mole of the compound, there are 2 moles of aluminum ions (Al3+). This means that the mole ratio of Al3+ to Al2(SO4)3 is 2:1.
Given that we have 0.42 mol of Al2(SO4)3, we can calculate the moles of Al3+ as follows:
Moles of Al3+ = 0.42 mol Al2(SO4)3 x (2 mol Al3+ / 1 mol Al2(SO4)3)
Moles of Al3+ = 0.42 mol Al2(SO4)3 x 2
Moles of Al3+ = 0.84 mol Al3+
Therefore, there are 0.84 moles of aluminum ions (Al3+) present in 0.42 mol of Al2(SO4)3.
It's important to note that the stoichiometry of the compound determines the mole ratio between the different species involved in the chemical formula. In this case, the 2:1 ratio of Al3+ to Al2(SO4)3 allows us to determine the number of moles of Al3+ based on the given amount of Al2(SO4)3.
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