Could you please help
How many kilojoules of energy are released when .1.60 kg of water cools from 80.0 ºC to 60.0 ºC?
Why do we use molecular models to represent atoms?
Combustion of 25.0 g of a hydrocarbon produces 86.5 g of co2. what is the empirical formula of the compound?
The empirical formula of the hydrocarbon is [tex]\boxed{{{\text{C}}_7}{{\text{H}}_5}}[/tex].
Further explanation:
Empirical formula:
It is atom’s simplest positive integer ratio in the compound. It may or may not be same as that of molecular formula. For example, empirical formula of sulfur dioxide is SO.
Combustion reactions:
These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.
Example of combustion reactions are as follows:
(a) [tex]{\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}[/tex]
(b) [tex]{{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+ 4{{\text{H}}_2}{\text{O}}[/tex]
[tex]{\text{C}}{{\text{O}}_2}[/tex] is formed as a product during combustion reactions.
Step 1: [tex]{\text{C}}{{\text{O}}_2}[/tex] is formed as a product during combustion reactions. Initially, we have to calculate the moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] . The formula to calculate moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is as follows:
[tex]{\text{Moles of C}}{{\text{O}}_2}=\dfrac{{{\text{Given mass of C}}{{\text{O}}_2}}}{{{\text{Molar mass of C}}{{\text{O}}_2}}}[/tex] ...... (1)
The given mass of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 86.5 g.
The molar mass of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 44 g/mol.
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Moles of C}}{{\text{O}}_2}&=\left( {{\text{86}}{\text{.5 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{44 g}}}}} \right)\\&={\text{1}}{\text{.9659 mol}}\\ &\approx{\text{1}}{\text{.966 mol}}\\\end{aligned}[/tex]
Step 2: During combustion, one mole of carbon reacts to form one mole of [tex]{\text{C}}{{\text{O}}_2}[/tex] .So the mass of C in the hydrocarbon is calculated as follows:
[tex]{\text{Mass of C}}=\left( {{\text{Moles of C}}{{\text{O}}_{\text{2}}}}\right)\left( {\dfrac{{{\text{Moles of C}}}}{{{\text{Moles of C}{{\text{O}}_{\text{2}}}}}}\right)\left( {{\text{Molar mass of C}}}\right)[/tex] ...... (2)
The moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 1.966 mol.
The molar mass of C is 12 g/mol.
The mole of C is 1 mol.
The moles of [tex]{\text{C}}{{\text{O}}_2}[/tex] is 1 mol.
Substitute these values in equation (2).
[tex]\begin{aligned}{\text{Mass of C}}&=\left( {{\text{1}}{\text{.966 mol}}} \right)\left( {\frac{{{\text{1 mol of C}}}}{{{\text{1 mol of C}}{{\text{O}}_{\text{2}}}}}} \right)\left( {\frac{{{\text{12 g}}}}{{{\text{1 mol}}}}} \right)\\&= {\text{23}}{\text{.592 g}}\\&\approx{\text{23}}{\text{.59 g}}\\\end{aligned}\\[/tex]
Step 3: Since the hydrocarbon consists of only carbon and hydrogen. The mass of hydrogen is calculated as follows:
[tex]{\text{Mass of H}}={\text{Mass of hydrocarbon}}-{\text{Mass of C}}[/tex] ...... (3)
The mass of hydrocarbon is 25 g.
The mass of carbon is 23.59 g.
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{Mass of H}}&={\text{25 g}}-{\text{23}}{\text{.59 g}}\\&= {\text{1}}{\text{.41 g}}\\\end{aligned}[/tex]
The formula to calculate moles of H is as follows:[tex]{\text{Moles of H}}=\dfrac{{{\text{Given mass of H}}}}{{{\text{Molar mass of H}}}}[/tex] ...... (4)
The given mass of H is 1.41 g.
The molar mass of H is 1.01 g/mol.
Substitute these values in equation (4).
[tex]\begin{aligned}{\text{Moles of H}}&=\left({{\text{1}}{\text{.41 g}}}\right)\left( {\frac{{{\text{1 mol}}}}{{{\text{1}}{\text{.01 g}}}}}\right)\\&={\text{1}}{\text{.396 mol}}\\ \end{aligned}[/tex]
The moles of carbon and hydrogen present in hydrocarbon are to be written with their corresponding subscripts. So the preliminary formula becomes,
[tex]{\text{Preliminary formula of hydrocarbon}}={{\text{C}}_{1.966}}{{\text{H}}_{1.396}}[/tex]
Step 4: Each of the subscripts is divided by the smallest subscript to get the empirical formula. In this case, the smallest one is 1.39. So the empirical formula of hydrocarbon is written as follows:
[tex]\begin{aligned}{\text{Empirical formula of hydrocarbon}}&={{\text{C}}_{\dfrac{{1.966}}{{1.396}}}}{{\text{H}}_{\dfrac{{1.396}}{{1.396}}}}\\&= {{\text{C}}_{1.408}}{{\text{H}}_1}\\\end{aligned}[/tex]
Step 5: Multiply each subscript of the empirical formula by 5, we get the final empirical formula as follows:
[tex]\begin{aligned}{\text{Empirical formula of hydrocarbon}}&={{\text{C}}_{5\left( {1.408} \right)}}{{\text{H}}_{5\left( 1 \right)}}\\&={{\text{C}}_7}{{\text{H}}_5}\\\end{aligned}[/tex]
Therefore, the empirical formula of hydrocarbon is [tex]{{\mathbf{C}}_{\mathbf{7}}}{{\mathbf{H}}_{\mathbf{5}}}[/tex] .
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Stoichiometry of formulas and equations
Keywords: empirical formula, C, H, C7H5, moles of CO2, C, H, 5, preliminary formula, whole number.
A hypothesis does_____ need to be correct in order for an experiment to be a success.
a salt is heated, evolves water, and turns from green to white. is the salt a hydrate
He henry's law constant of oxygen in water at 25 °c is 773 atm mol-1kg h2o. calculate the molality of oxygen in water under a partial pressure of 0.20 at
To calculate the molality of oxygen in water under a partial pressure of 0.20 atm, apply Henry's law with the given constant (773 atm mol⁻¹kg H₂O), resulting in a molality of 154.6 mol kg⁻¹.
The question asks to calculate the molality of oxygen in water at a specific temperature (25 °C) and partial pressure, using the given Henry's law constant. To find the molality, we need to apply Henry's law, which states that the concentration of a dissolved gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.
Hence, the formula for Henry's law is:
C = kP
where:
C is the molar concentration of the dissolved gas (molality in this case),
k is the Henry's law constant,
P is the partial pressure of the gas above the liquid.
Using the given Henry's law constant for oxygen in water (773 atm mol⁻¹kg H₂O) and the partial pressure of oxygen (0.20 atm), we can solve for C as follows:
C = kP
C = 773 atm mol⁻¹kg H₂O × 0.20 atm
C = 154.6 mol kg⁻¹
So, the molality of oxygen in water under a partial pressure of 0.20 atm is 154.6 mol kg⁻¹.
A sample of pure water has a hydronium concentration of 1.0 × 10-7 M. What is the pH of the water?
Answer:The pH of the water is 7.
Explanation;
The pH of the solution is defined as negative logarithm of [tex]H^+[/tex] or hydronium ions ions in the solution.
[tex]pH=-\log[H^+][/tex]
So, [tex]H^+[/tex] concentration of water = [tex]1.0\times 10^{-7} m[/tex]
[tex]pH=-\log[1.0\times 10^{-7}]=7[/tex]
The pH of the water is 7.
When chlorine and magnesium react chemically, how many chlorine atoms will combine with one magnesium atom?
Write the chemical formula of the compound that is formed
Consider the resonance structures of formate. the first lewis structure of formate has a central carbon atom. a hydrogen atom and two osygen atoms are bonded to the carbon atom. the bond between carbon and hydrogen is a single bond. one of the bonds between carbon and oxygen is a single bond and the other bond is a double bond. the single bonded oxygen has three lone pairs of electrons and a negative one charge. the double bonded oxygen has two lone pairs of electrons. the second lewis structure of formate has all of the same atom connectivities, but the double and single bonds between oxygen and carbon are switched from the first strucutre. select the true statements about the resonance structures. the actual structure of formate switches back and forth between the two resonance forms. each carbon–oxygen bond is somewhere between a single and double bond. each oxygen atom has a double bond 50% of the time. the actual structure of formate is an average of the two resonance forms.
The statement "each carbon-oxygen bond is somewhere between a single and double bond" and "the actual structure of formate is an average of the two resonance forms" are correct. The resonance structures represent an average distribution of electrons across all valid structures, not constant flipping between these structures.
Explanation:In regards to the resonance structures of formate, the statements "each carbon–oxygen bond is somewhere between a single and double bond" and "the actual structure of formate is an average of the two resonance forms" are both true.
Resonance describes the situation where more than one valid Lewis structure can be drawn for a particular molecule. The resonance structure is not a rapid equilibrium between the structures but rather an average of the different possible structures, called resonance forms.
In the case of formate, which has two resonance forms, the molecule doesn't constantly flip between these two structures. Instead, the electrons are distributed in a way that is an average of these two resonance forms. This is why each carbon-oxygen bond in formate is described as being somewhere between a single and a double bond, as the characteristics of the bond are shared across both resonance forms.
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Which two elements have the most similar chemical properties? 1. beryllium and magnesium 2. hydrogen and helium 3. phosphorus and sulfur 4. potassium and strontium
beryllium and magnesium is your answer
The two elements which have the most similar chemical properties are: Beryllium and Magnesium.
Discussion:
The two elements are present in the Alkali earth metals group of the periodic table.
They are both characterized by the possession of 2 Valence electrons on their outermost shell and as such are able to undergo similar chemical reactions.
Ultimately, beryllium and magnesium have the most similar chemical properties.
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What is the density (g/mL) of an object that has a mass of 0.03 kg and occupies a volume of 25 mL?
Answer : The density of an object is 1.2 g/mL
Explanation :
Density : It is defined as the mass contained per unit volume.
Formula used :
[tex]Density=\frac{Mass}{Volume}[/tex]
Given :
Mass of object = 0.03 kg = 30 g
conversion used : (1 kg = 1000 g)
Volume occupied by object = 25 mL
Now put all the given values in the above formula, we get :
[tex]Density=\frac{30g}{25mL}=1.2g/mL[/tex]
Therefore, the density of an object is 1.2 g/mL
11. Which of the following accurately describes properties of valence? A. Nonmetallic elements tend to have a positive valence and tend to be electron borrowers. B. The smaller the number of electrons an atom has to borrow or to lend, the greater the activity of the atom. C. Metals tend to have a negative valence and tend to be electron borrowers. D. The greater the number of electrons an atom has to borrow or to lend, the greater the activity of the atom.
Calculate the number of Li atoms in 5.1 moles of Li
The number of Li atoms in 5.1 moles of Li is 3.07 x 10²⁴ atoms.
What are atoms?Atoms are defined as the smallest piece of matter that can be separated without sending electrically charged particles flying.
It can also be defined as the smallest piece that carries an element's characteristics. Subatomic particles, which make up an atom, are uncreatable.
There are various types of atoms.
DescriptionStableIsotopesRadioactive IonsAntimatterIn a chemical reaction, atoms cannot be formed or destroyed since they are indivisible units. The mass and chemical characteristics of each atom of a specific element are the same. Different elements' atoms have varying weight and chemical characteristics. Compounds are created when atoms combine in ratios of small whole numbers.
Moles of Li = 5.1 x 6.022 x 10²³
= 3.07 x 10²⁴ atoms.
Thus, the number of Li atoms in 5.1 moles of Li is 3.07 x 10²⁴ atoms.
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What is the most common type of climate/ecosystem found in the Congo River basin?
Explain how the determination of the rate law equation significantly differs from the determination of the equilibrium constant keq expression.
88.5 mol of P4O10 contains how many moles of P?
Answer : The number of moles of phosphorous are 354 moles.
Explanation :
The formula of given compound is, [tex]P_4O_{10}[/tex]
In [tex]P_4O_{10}[/tex] compound, there 4 moles of phosphorus and 10 moles of oxygen.
As we are given that the moles of [tex]P_4O_{10}[/tex] is 88.5 mole. Now we have to determine the number of moles of phosphorous (P).
As, 1 mole of [tex]P_4O_{10}[/tex] has 4 moles of phosphorous
So, 88.5 mole of [tex]P_4O_{10}[/tex] has [tex]4\times 88.5=354moles[/tex] of phosphorous
Therefore, the number of moles of phosphorous are 354 moles.
88.5 mol of [tex]P_4O_{10}[/tex] contains 354 moles of P. A mole is a unit of measurement used in chemistry to represent how much of a substance is present.
In chemistry, a mole is a unit of measurement that is used to express how much a material is present. It is described as the quantity of a substance that has exactly the same number of atoms, molecules, or ions as there are in exactly 12 grammes of pure carbon-12. The mole idea is essential to chemistry because it enables researchers to connect a substance's mass to its particle count. The idea of molar mass, or the mass of one mole of a substance, is used to describe this relationship. The unit of molecular mass is grammes per mole (g/mol).
88.5 mol [tex]P_4O_{10}[/tex] × (4 mol P / 1 mol P4O10) = 354 mol P
88.5 mol of [tex]P_4O_{10}[/tex] contains 354 moles of P
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Find the number of moles of water that can be formed if you have 210 mol of hydrogen gas and 100 mol of oxygen gas.
Answers section 4.2 structure of the nuclear atom 1. a sulfur-32 atom contains 16 protons, 16 neutrons, and 16 electrons. what is the mass (in grams) of a sulfur-32 atom? 2. the mass of a neutron is 1.67 x 10-24 g. approximately what number of neutrons would equal a mass of one gram?
The mass of a sulfur-32 atom is approximately 8.01 x 10^-23g. There are approximately 5.97 x 10^23 neutrons in one gram.
Explanation:The mass of a sulfur-32 atom, which is a type of nuclear atom, can be determined by summing the masses of its protons, neutrons, and electrons. Since a sulfur-32 atom has 16 protons, 16 neutrons, and 16 electrons, and given that the mass of a proton or neutron is approximately 1.67 x 10^-24 g, the total mass of the atom would be 48 x 1.67 x 10^-24g, which equals approximately 8.01 x 10^-23g.
Furthermore, to determine the number of neutrons that would equal a mass of one gram, we divide one gram by the mass of a single neutron. This results in approximately 5.97 x 10^23 neutrons.
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The process that makes ice cubes shrink as they sit in a freezer is called:
A- Sublimation
B- Condensation
C- Freezing
D- Boiling
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Instructions:Select ALL the correct answers.
Which of these factors make hydrogen fuel cells a better option than burning fossil fuels?
Hydrogen fuel cells have a higher energy efficiency.
Hydrogen fuel cells create less pollution.
Burning fossil fuels relies on outdated devices and technology.
Pure hydrogen is more readily available than fossil fuels.
Hydrogen fuel cells are more expensive than fossil fuels.
Answer:
Hydrogen fuel cells create less pollution.
Burning fossil fuels relies on outdated devices and technology.
Hydrogen fuel cells have a higher energy efficiency.
Explanation:
Hello,
The first option - Hydrogen fuel cells create less pollution - is selected due to the fact that fuel cell consists of an electrolyte sandwiched between two electrodes, an anode and a cathode, so electrical energy is used instead of the burning of a fossil fuel which gives off pollutant gases such as carbon, nitrogen and sulfur oxides.
The second option - Burning fossil fuels relies on outdated devices and technology - is selected due to even when new technologies are being developed to enhance the extraction of energy from the combustion of fossil fuels, they are based on the same idea, combustion which contributes to pollution.
The third option - Hydrogen fuel cells have a higher energy efficiency - is selected because a typical combustion process converts the 35% of the total fuel into usable energy meanwhile the hydrogen fuel cells reach the up to the 60% or more.
Best regards.
How many chloride ions are present in 0.100 mol of MgCl2
To identify the amount of chloride ions present in 0.1 mol of MgCl2, first know the amount of mols of Chloride. The answer is 0.2 mol since you just need to multiply 0.1 to 2 because there are 2 chloride in the compound. Then, multiply the mols to 6.02x10^23. The answer would be 1.204x10^23 chloride ions.
Answer : The number of chloride ions present in 0.100 mole of [tex]MgCl_2[/tex] are, [tex]1.2044\times 10^{23}[/tex]
Explanation : Given,
Moles of [tex]MgCl_2[/tex] = 0.100 mole
As we know that, 1 mole contains [tex]6.022\times 10^{23}[/tex] number of ions.
In [tex]MgCl_2[/tex], there are one magnesium ion and two chloride ions.
As, 1 mole [tex]MgCl_2[/tex] contains [tex]2\times (6.022\times 10^{23})[/tex] number of chloride ions.
So, 0.100 mole [tex]MgCl_2[/tex] contains [tex]0.100\times 2\times (6.022\times 10^{23})=1.2044\times 10^{23}[/tex] number of chloride ions.
Therefore, the number of chloride ions present in 0.100 mole of [tex]MgCl_2[/tex] are, [tex]1.2044\times 10^{23}[/tex]
Determine the concentrations of hg22 and cl– in a saturated solution of hg2cl2 in
a.pure water (assume μ = 0).
To solve this, we need the solubility constant for Hg2Cl2, which is:
Ksp = 1.4 x 10^-18
The ionic formula would be:
Hg2Cl2(s) ---> Hg2 2+(aq) + 2Cl-(aq)
Therefore the total amounts of ions produced = x M Hg2 2+ and 2x M
Cl-
The formula for Ksp in this case is:
Ksp = [Hg2 2+] [Cl-]^2
1.4 x 10^-18 = (x) * (2x)^2
1.4 x 10^-18 = 4x^3
x = 7.0 x 10^-7 M = [Hg2 2+]
2x = 14 x 10^-7 M = [Cl-]
What mass of znco3 contains 6.11×1022 o atoms? give your answer correctly to three significant digits?
6.11×10²² oxygen atoms are present in 4.24 grams of ZnCO₃. The molar mass of ZnCO₃ is 125.34 g/mol.
Given information:
The molecular weight of ZnCO₃ = 125.34 g/mol
Number of oxygen atoms = 6.11×10²²
1 mole of ZnCO₃ → 3 moles of oxygen atoms
1 mole of ZnCO₃ → atoms
125.34 g/mol ZnCO₃ → 3 × 6.023 × 10²³ atoms
Mass is a measure of the amount of matter in an object. It is a fundamental property of matter, and it does not change with the object's location or motion. The SI unit of mass is the kilogram (kg).
Now, the mass of ZnCO₃ is:
[tex]\rm Mass = \frac{6.11 \times 10^{22} \times 125.4}{3 \times 6.023 \times 10^{23}}\\\rm Mass = \frac{6.11 \times 10^{22} \times 125.4\times 10^{-23}}{3 \times 6.023 }\\Mass = \frac{766.194}{18.069}\times10^{-1}\\Mass = \frac{76.62}{18.069}\\[/tex]
Mass = 4.24 grams
Therefore, the mass of ZnCO₃ is 4.24 grams which contains 6.11×10²² oxygen atoms.
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To find the mass of ZnCO3 that contains 6.11×10^22 O atoms, we can calculate the molar mass of ZnCO3, use it to convert the number of oxygen atoms into moles, and then calculate the mass of ZnCO3. The mass of ZnCO3 is 12.73 g.
Explanation:To find the mass of ZnCO3 that contains 6.11×10^22 O atoms, we need to calculate the molar mass of ZnCO3 and then use it to convert the number of oxygen atoms into moles. Finally, we can use the molar mass of ZnCO3 to calculate the mass of the compound.
The molar mass of ZnCO3 is calculated by summing the atomic masses of its constituent elements: Zn (65.38 g/mol), C (12.01 g/mol), and 3 O atoms (16.00 g/mol).
Molar mass of ZnCO3 = 65.38 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = 125.38 g/mol
Now, let's calculate the number of moles of O atoms:
Number of moles of O atoms = (6.11×10^22) / (6.022×10^23)
= 0.1015 mol
Finally, we can calculate the mass of ZnCO3:
Mass of ZnCO3 = (0.1015 mol) * (125.38 g/mol)
= 12.73 g
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A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what is the final temperature of both substances at thermal equilibrium?
The final temperature at thermal equilibrium can be calculated using the concept of conservation of energy and the specific heat capacities of gold and water.
Explanation:To find the final temperature of the gold and water system when they reach thermal equilibrium, we need to apply the concept of conservation of energy. This concept suggests that in an isolated system, the heat lost by the hot object (the gold) will be equal to the heat gained by the cold object (the water). Since the system is at equilibrium, the heat lost is equal to the heat gained, hence the formula: Cgold × mgold × (Tinitial, gold - Tfinal) = -Cwater × mwater × (Tfinal - Tinitial, water), where Cgold and Cwater are the specific heat capacities of gold and water, T is the temperature and m is the mass.
We also need to know the specific heat capacities of gold and water. The specific heat capacity of gold is 0.129 J/g °C and for water, it's 4.18 J/g °C. Substituting those values along with the original temperatures and masses, we can solve for the final temperature, Tfinal.
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FeCl2 Express your answers as ions separated by a comma.
The ionic compound formed from Fe3+ and Cl- is FeCl3, called iron(III) chloride. For dissociation in water, FeCl2 separates into Fe2+ and Cl-. The complete ionic equation for the reaction between FeCl2 and AgNO3 includes all ions present with AgCl precipitating out, and the net ionic equation shows just the ions involved in the formation of the precipitate.
Naming the Ionic Compound Formed from Fe3+ and Each Anion
To name the ionic compound formed from Fe3+ and an anion, you follow a set of rules. Firstly, you write the name of the cation including the charge in roman numerals enclosed in small brackets, followed by the name of the anion. For Fe3+, we use the name Iron(III), and for a chloride ion (Cl-), the name is chloride. The compound's formula is FeCl3, which is called iron(III) chloride. Remember that the subscript in the chemical formula indicates the number of anions needed to balance the charge of the cation.
Compound Dissociation in Water
To write a balanced equation for how an ionic compound like FeCl2 dissociates in water, you would represent it as follows:
FeCl2(s) -> Fe2+(aq) + 2 Cl-(aq)
This shows the separation of the ionic compound into its individual ions when dissolved in water.
Complete Ionic Equation for FeCl2 and AgNO3
The complete ionic equation for the reaction between FeCl2(aq) and AgNO3(aq), consulting solubility rules, would look like this:
Fe2+(aq) + 2 Cl-(aq) + 2 Ag+(aq) + 2 NO3-(aq) -> 2 AgCl(s) + Fe2+(aq) + 2 NO3-(aq)
Net Ionic Equation for FeCl2 and AgNO3
The net ionic equation, after removing the spectator ions, for the same reaction would be:
2 Cl-(aq) + 2 Ag+(aq) -> 2 AgCl(s)
Which of the following is a heterogeneous mixture?
A) vinegar and water
B) milk
C) Oil and vinegar
D) Air
A 35.0-ml sample of 1.00 m kbr and a 60.0-ml sample of 0.600 m kbr are mixed. the solution is then heated to evaporate water until the total volume is 50.0 ml. how many grams of silver nitrate are required to precipitate out silver bromide in the final solution?
To precipitate all the bromide ions present in the final solution of KBr, 12.061g of silver nitrate is needed. The calculation is based on molarity, volume and reaction stoichiometry.
Explanation:The first step to finding the answer is to calculate the amount of KBr in each solution. The amount of solute in a solution is given by the formula: Volume (L) × Molarity (M). So for the 35.0 mL sample of 1.00 M KBr, that would be 0.035 L × 1.00 mol/L = 0.035 moles of KBr. For the 60.0 mL sample of 0.600 M KBr, the calculation is 0.060 L × 0.600 mol/L = 0.036 moles of KBr. Thus, a total of 0.035 + 0.036 = 0.071 moles KBr is present in the final solution.
The reaction between KBr and silver nitrate (AgNO3) is a one-to-one reaction: KBr + AgNO3 → AgBr + KNO3. Therefore, 0.071 moles of AgNO3 are required to precipitate all the bromide ions. The molar mass of AgNO3 is approximately 169.87 g/mol, therefore the total mass of AgNO3 needed is 0.071 moles × 169.87 g/mol = 12.061 g.
Hence, 12.061 g of silver nitrate are required to precipitate out silver bromide in the final solution.
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The amount of silver nitrate required is approximately 10.53 grams. This is calculated by first determining the total moles of KBr in the solution and then using the 1:1 stoichiometry of the reaction to find the equivalent moles (and thus mass) of AgNO3 required.
Explanation:This is a stoichiometry problem that involves determining the amount of silver nitrate needed to precipitate out silver bromide in a solution. First, we need to determine the number of moles of KBr in the final solution. The original solutions have volumes of 35.0 ml and 60.0 ml and molarities of 1.00 M and 0.600 M respectively, so the number of moles of KBr is (35.0 ml x 1.00 mol/L) + (60.0 ml x 0.600 mol/L) = 0.062 mol. After evacuating some water, the volume decreases to 50.0 ml.
Now, since silver nitrate and potassium bromide react in a 1:1 ratio to form silver bromide, the moles of silver nitrate needed will be equal to the moles of KBr. Thus, we need 0.062 moles of AgNO3. The molecular weight of AgNO3 is 169.87 g/mol, so the mass of AgNO3 needed is (0.062 mol) x (169.87 g/mol) = 10.53 g. So, you will need about 10.53 grams of silver nitrate to precipitate out silver bromide in this solution.
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There are ________ mol of bromide ions in 0.500 l of a 0.400 m solution of albr3.
The four isotopes of lead are shown below, each with its percent by mass abundance and the composition of its nucleus. using these data, calculate the approximate atomic mass of lead.
The average atomic mass of lead, will be approximately 207.2 g/mol.
To calculate the approximate atomic mass of lead using the isotopic masses and their relative abundances. The atomic mass is a weighted average based on the abundances of each isotope.
Identify the element using the atomic number (number of protons).Calculate the mass number (A) of each isotope by adding the number of protons to the number of neutrons.Write the symbol of each isotope, with the mass number as a superscript and the number of protons as a subscript, to the left of the element symbol.Multiply the mass number of each isotope by its percentage abundance and then sum the products.Divide the total by 100 to find the average atomic mass.For lead (Pb), the calculation using given isotopes would look like this:
(1.40% x 203.973) + (24.10% x 205.974) + (22.10% x 206.976) + (52.40% x 207.977)Sum of all products divided by 100 gives the approximate atomic mass.Thus, the approximate atomic mass of Pb is 207.2 g/mol.
The complete question is:
The four isotopes of lead are shown below, each with its percent by mass abundance and the composition of its nucleus. using these data, calculate the approximate atomic mass of lead. 82p 122n 1.37% 82p 124n 26.26% 125n 20.82 82p 126n 51.55% 24 Mass # 81 37 6.