a.
The balanced equation for the reaction between sulfuric acid and aluminium hydroxide is,
3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O
The products formed from the reaction between
aluminium hydroxide and sulfuric acid are Al₂(SO₄)₃ and H₂O
The limiting reactant is H₂SO₄
The stoichiometric ratio between H₂SO₄
and Al₂(SO₄)₃ is 3 : 1
Reacted moles of H₂SO₄ =
0.306 mol
Hence the moles of Al₂(SO₄)₃ formed = 0.306 mol / 3
= 0.102 mol
Molar mass of Al₂(SO₄)₃ = 342 g/mol
Mass of Al₂(SO₄)₃ formed = 0.102 mol x 342 g/mol
= 34.884 g
The stoichiometric ratio between H₂SO₄
and H₂O is 3 : 6
Reacted moles of H₂SO₄ =
0.306 mol
Hence the moles of H₂O formed = 0.306 mol x (6 / 3)
= 0.612 mol
Molar mass of H₂O = 18 g/mol
Mass of H₂O formed = 0.612 mol x 18 g/mol
= 11.016 g
The limiting reactant is sulfuric acid. The mass of excess reactant remaining is 13.075 g. The mass of aluminum sulfate formed is 104.738 g and the mass of water formed is 5.562 g.
Explanation:The question is about a chemical reaction where sulfuric acid reacts with aluminum hydroxide by double displacement to produce aluminum sulfate and water.
To identify the limiting reactant, we need to know the moles of sulfuric acid and aluminum hydroxide. Using the molecular weights (98.09 g/mol for sulfuric acid and 78.0 g/mol for aluminum hydroxide), we find that we have 0.306 moles of sulfuric acid and 0.32 moles of aluminum hydroxide. The reaction ratio from the balanced chemical equation is 1:2, implying we need twice as many moles of sulfuric acid as aluminum hydroxide. Therefore, sulfuric acid is the limiting reactant.Using the stoichiometry of the reaction, we find out that all the sulfuric acid is consumed, leaving excess aluminum hydroxide. The mass of excess reactant remaining is calculated by subtracting the mass of aluminum hydroxide consumed from the initial mass of aluminum hydroxide. Aluminum hydroxide consumed is (0.306 moles / 2) * 78.0 g/mol = 11.925 g. So, the mass of excess reactant remaining is 25.0 g (initial) - 11.925 g(consumed) = 13.075 gIn terms of mass of each product formed, we get aluminum sulfate and water. The mass of aluminum sulfate formed is the moles of limiting reactant times the molar mass of the product, i.e., 0.306 moles * 342.15 g/mol = 104.738 g. The mass of water formed is also calculated similarly, giving 5.562 g.Learn more about Limiting Reactant here:https://brainly.com/question/33417913
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Iron, an important component of the steel used in the construction of buildings, bridges, and railroads, combines with oxygen to form fe2o3, which we recognize as rust. at room temperature, iron
Which is an acceptable Lewis structure for a diatomic nitrogen molecule?
The Lewis structure for a diatomic nitrogen molecule would indicate a triple bond between the two nitrogen atoms, with each atom having a pair of lone electrons. The concept of resonance doesn't apply to this molecule.
Explanation:The diatomic nitrogen molecule has a triple bond between the two nitrogen atoms. A Lewis structure, in this context, represents the arrangement of electrons in a molecule. The Lewis structure for N2 would therefore show the nitrogen atoms sharing three pairs of electrons - indicating the triple bond - with each nitrogen atom also having one pair of lone (non-bonding) electrons.
Using the concepts of resonance and electron-pair geometries, it's possible to show that the actual distribution of electrons is an average between various possible Lewis structures. However, the resonance only applies to molecules where it's not possible to depict the molecule's structure using a single Lewis structure, as in the case of nitrogen dioxide (NO2) and related compounds, not diatomic nitrogen.
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What will be the effect on the amount of gas produced if the experiment is repeated using 0.35g of k?
The effect on the amount of gas produced if the experiment is repeated using 0.35g of potassium is that less amount of hydrogen gas is produced.
What is the stoichiometry?Stoichiometry of any reaction gives idea about the amount of species present and before the completion of the reaction.
Chemical reaction with repect to Lithium (Li) and Potassium (K) will be represented as:
2K + 3H₂O → 2H₂ + 2KOH
2Li + 3H₂O → 2H₂ + 2LiOH
From the stoichiometry of the reaction it is clear that 2 moles of K & Li produces 2 moles of hydrogen gas.
Moles will be calculated as:
n = W/M, where
W = given mass = 0.35g
M = molar mass
Moles of 0.35g Li = 0.35 / 6.9 = 0.0507 moles
Moles of 0.35g K = 0.35 / 39 = 0.0089 moles
From the moles of both elements it is clear that lithium will produce more gas.
Hence on replacing lithium with potassium, less gas will be produced.
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A 5.00 L sample of helium at STP expands to 20.0 L. What is the new pressure on the gas?
Le Châtelier's principle tells us that chemical _____ will adjust in an attempt to remove a stressor.
a. formulas
b. equations
c. equilibrium
d. concentrations
What is the most abundant gas in our atmosphere?
1. oxygen
2. hydrogen
3. nitrogen
4. water
What will most likely happen if a sound wave moves from the air through a solid?
Answer:
D
Explanation:
YOUR WELCOME
A gas is contained in a thick walled balloon. when the pressure changes from 417 mm hg to 576 mm hg, the volume changes from ____ L to 4.78 L and the temperature changes from 497 K to 386 K
Answer: 8.5
Explanation:
Calculate h3o+ in an aqueous solution that is 0.10M NaNO2 and 0.055M HNO2
Differences in which property allows the separation of a sample of sand and sea water by filteration
Filtration is a type of process in which components of a mixture are separated based in the size. The sea water and sand can be separated by filtration due to different particle size.
Seawater is a mixture of several substances, including salt, sand, and stones. The substances of seawater can be separated via filtration. The filter paper in this technique separates the substances based on the size of particles.
The filtration separates sand and seawater as:
Seawater and sand are dissolved in each other. When the filtration technique is performed, it separates out the water from sand. The size of the sand particles are larger than the pore size of filter paper, thus, it easily separates the sand from seawater.Therefore, difference in the particle size allows the separation of seawater and sand.
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Which reaction may be described as an acid-base neutralization reaction? 1. be(oh)2 + h2so4 → beso4 + 2 h2o 2. ba(oh)2 + 2 licl → bacl2 + 2 lioh 3. naoh + h2o → h3o + + nao− 4. hno2 + h2o → h3o + + no− 2?
In the third period of the periodic table sodium is followed by magnesium aluminum silicon and phosphorus which of these elements has the smallest atomic size ?
What mass of KBr is present in 25 mL of a 0.85 M solution of potassium chloride?
Answer : The correct answer for mass of KBr = 2.53 g
Given :
Molarity of KBr solution = 0.85 M
Volume of KBr solution = 25 mL
Converting volume from mL to L ( 1 L = 1000 mL )
[tex] Volume of solution = 25 mL * \frac{1 L }{1000mL} [/tex]
Volume of solution = 0.025 L
Mass of KBr = ?
Mass of KBr can be calculated using following steps :
1) To find mole of Kbr :
Mole of KBr can be calculated using molarity .
Molarity : It is defined as mole of solute present in volume of solution in Liter .
It uses unit as M or [tex] \frac{mol}{L} [/tex]
It can be expressed as :
[tex] Molarity = \frac{mol of solute (mol)}{volume of solution (L)} [/tex]
Plugging value of molarity and volume
[tex] 0.85 \frac{mol}{L} = \frac{mol of Kbr}{0.025 L} [/tex]
Multiplying both side by 0.025 L
[tex] 0.85 \frac{mol}{L} * 0.025 L = \frac{mole of KBr}{0.025 L} * 0.025 L [/tex]
Mole of KBr = 0.02125
2) To find mass of Kbr :
Mass of Kbr can be calculated using mole . Mole can be expressed as :
[tex] Mole (mol) = \frac{mass (g) }{molar mass \frac{g}{mol} } [/tex]
Mole of Kbr = 0.02125 mol
Molar mass of KBr = 119.00 [tex] \frac{g}{mol} [/tex]
Plugging values in mole formula
[tex] 0.02125 mol = \frac{mass (g)}{119.00 \frac{g}{mol}} [/tex]
Multiplying both side by 119.00 [tex] \frac{g}{mol} [/tex]
[tex] 0.02125 mol * 119.00 \frac{g}{mol} = \frac{mass (g)}{119.00 \frac{g}{mol}} * 119.00\frac{g}{mol} [/tex]
Mass of KBr = 2.53 g
Final answer:
To find the new volume of a diluted 0.885 M KBr solution after dilution to 0.500 M, use the formula (M1×V1) / M2, resulting in a new volume of 266.85 mL.
Explanation:
The subject of this question is determining the new volume of a diluted potassium bromide (KBr) solution in chemistry. When a 0.885 M solution of KBr with an initial volume of 76.5 mL is diluted to a concentration of 0.500 M, we can use the concept of molarity (M), which is defined as moles of solute per liter of solution, to find the new volume.
To find the new volume (V2) after dilution, we can apply the formula M1×V1 = M2×V2, where M1 and V1 are the initial molarity and volume, and M2 is the final molarity. By rearranging the formula, V2 = (M1×V1) / M2. Substituting the given values, we get V2 = (0.885 moles/L × 76.5 mL) / 0.500 moles/L = 133.425 mL / 0.500 = 266.85 mL. Thus, the new volume is 266.85 mL.
In the reaction 2co(g) + o2(g) â 2co2(g), what is the ratio of moles of oxygen used to moles of co2 produced? select one:
A substance that is not soluble in a polar solvent is
explain why many compounds that contain one or more hydrogen atoms are not classified as arrhenius acids
Final answer:
Not all compounds containing hydrogen atoms are classified as Arrhenius acids because they do not release hydrogen ions when dissolved in water. Methane and acetic acid are examples of compounds that contain hydrogen atoms but are not classified as Arrhenius acids.
Explanation:
In order for a compound to be classified as an Arrhenius acid, it must release hydrogen ions (H+) when dissolved in water. However, not all compounds that contain hydrogen atoms are classified as Arrhenius acids because they do not release H+ ions. For example, methane (CH4) and acetic acid (CH3COOH) both contain hydrogen atoms but they are not classified as Arrhenius acids because their hydrogen atoms are not capable of ionizing.
If a sample of a certain solution is determined to have a [h3o+][h3o+] concentration of 4.91×10−44.91×10−4 moles/liter, what is its ph? round off your answer to one decimal place.
Which phrase describes an al atom?
a.a negatively charged nucleus, surrounded by negatively charged electrons
b.a negatively charged nucleus, surrounded by positively charged electrons?
An atom is defined as the microscopic particle having positively charged nucleus around which negatively charged electrons revolve. In aluminum atom, none of the options give are correct.
Aluminum is an atom made up of 13 electrons and 13 protons. In an atom, there are several sub-particles, which are:
Protons are the positively charged sub-particle that are present inside the nucleus.Electrons are the negatively charged sub-particles that revolves around the nucleus. Neutrons does not have any charge, and is found in the nuclei of atoms.Therefore, the correct statement is that aluminum atom have positively charged nucleus around which electrons revolve. Hence, the given options are incorrect.
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calculate the pressure in atm, if 0.00825 moles occupies 174 mL at -15 celcius.
T = 15°C = 288.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p =
n·R·T / V.
p = 0.00825 mol · 0.08206 L·atm/mol·K · 288.15 K /
0.174 L.
p = 1.12 atm.
The pressure is calculated using the Ideal Gas Law. Given the moles, volume in liters, and temperature in Kelvin, the pressure is approximately 0.100 atm. This calculation involves converting units and applying the gas constant.
To calculate the pressure in atmospheres (atm) given the number of moles, volume, and temperature, we need to use the Ideal Gas Law formula:
PV = nRT
Here:
P is the pressure in atm.V is the volume in liters. Since the given volume is in milliliters (174 mL), we need to convert it to liters:174 mL × 0.001 L/mL = 0.174 L.
n is the number of moles, which is 0.00825 moles.R is the ideal gas constant, which is 0.0821 (L atm)/(K mol).T is the temperature in Kelvin. We need to convert the temperature from Celsius to Kelvin:T = -15 °C + 273 = 258 K.
Now, we substitute the known values into the Ideal Gas Law equation:
P × 0.174 L = 0.00825 moles × 0.0821 (L atm)/(K mol) × 258 K
Simplify and solve for P:
P = (0.00825 moles × 0.0821 × 258) / 0.174
P ≈ 0.100 atm
A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator and titrates the acid to neutrality using 43.7 mL of 0.13 M NaOH base.
a. Write and balance the neutralization reaction of the acid and base
b. What is the molarity of the acid?
How many kj of heat are needed to completely vaporize 1.30 moles of h2o? the heat of vaporization for water at the boiling point is 40.6 kj/mole?
Explanation:
Heat of vaporization is defined as the amount of heat required to change one mole of a liquid into vapor state without any change in the temperature.
It is known that for 1 mole of water, latent heat of vaporization is 40.6 kJ/mol.
Therefore, heat of vaporization for 1.30 moles will be calculated as follows.
[tex]1.30 moles \times 40.6 kJ/mol[/tex]
= 52.78 kJ
Thus, we can conclude that the 52.78 kJ of heat are needed to completely vaporize 1.30 moles of [tex]H_{2}O[/tex].
Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses
The molar mass of a gas influences its diffusion and effusion rates, according to Graham's law of effusion. Lighter gas molecules diffuse and effuse faster than heavier ones due to their movement speed. The ratios of diffusion and effusion rates, both dependent on the molar mass, remain constant.
Explanation:The molar mass of a gas significantly affects its diffusion and effusion rates due to the principles identified in Graham's law of effusion. According to this law, the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. Hence, lighter gas molecules will effuse and diffuse at a faster rate as compared to heavier ones.
Furthermore, the average speed of gas molecules impacts the rate of effusion. As a consequence, heavier gas molecules, which move more slowly due to their larger mass, will diffuse and effuse slower compared to lighter molecules. This is because it is more challenging for heavier molecules to move between the molecules of the substance they are moving through.
Lastly, diffusion and effusion rates both depend on the molar mass of the gas involved. Although their rates are not identical, the ratios of their rates remain consistent. This is another way the molar mass of a gas influences its diffusion and effusion rates.
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What can the size and shape of a sand dune tell you about how it was formed?
Help Please earth science
Use the image to determine which of the following statements is true?
1) The granite is younger than unit B.2) Unit B and the granite are the same age.3) The relative ages of teh granite and unit B cannot be determined from the information given.4) Unit B is younger than the granite.The energy stored in the motion or vibration of molecules in a substance is a definition for
A. heat.
B. work.
C. radioactivity.
D. electromagnetism.
Assume that the rate law for a generic chemical reaction is rate = [a][b]3 . what is the reaction order in a, the reaction order in b, and the overall reaction order?
The order of reaction in a is 1 while the order of reaction in b is 3.
What is order of reaction?The rate law is obtained from the reaction equation or from experiment. We can determine the order of reaction by summing up the exponents in the rate law as shown in the equation of the rate law.
Now, we have; rate = k[a][b]3, the order of reaction in a is 1 while the order of reaction in b is 3.
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Which must be the same when comparing 1 mol of oxygen, O2, with 1 mol of carbon monoxide has, CO?
An element found in another galaxy exists as two isotopes. if 80.0 percent of the atoms have an atomic mass of 80.00 atomic mass units and the other 20.0 percent have an atomic mass of 82.00 atomic mass units, what is the atomic mass of the element?
"what is the concentration of positive charge and mass in rutherford's atomic model called?"
When bismuth-212 undergoes alpha decay, it becomes which of the following?
A asatine-208 + an alpha particle
B lead-210 + an alpha particle
C polonium-210 + and alpha particle
D thallium-208 + and alpha particle