1. The mass (in grams) of FeCl₂ contained in the sample is 14.96 grams
2. The mass (in grams) of KBr present in the solution is 16.66 grams
How do i determine the mass present?1. The mass of FeCl₂ contained in the sample can be obtained as shown below:
First, we shall obtain the mole
Volume = 155 mL = 155 / 1000 = 0.155 LMolarity = 0.762 MMole of FeCl₂ =?Mole = molarity × volume
Mole of FeCl₂ = 0.762 × 0.155
Mole of FeCl₂ = 0.118 mole
Finally, we shall determine the mass of FeCl₂ present in the sample. Details below:
Mole of FeCl₂ = 0.118 moleMolar mass of FeCl₂ = 126.75 g/molMass of FeCl₂ = ?Mass = Mole × molar mass
Mass of FeCl₂ = 0.118 × 126.75
Mass of FeCl₂ = 14.96 grams
2. The mass of KBr present in the solution can be obtained as shown below:
First, we shall obtain the mole
Volume = 0.4 LMolarity = 0.350 MMole of KBr =?Mole = molarity × volume
Mole of KBr = 0.350 × 0.4
Mole of KBr = 0.14 mole
Finally, we shall determine the mass of KBr in the solution. Details below:
Mole of KBr = 0.14 moleMolar mass of KBr = 119 g/molMass of KBr = ?Mass = Mole × molar mass
Mass of KBr = 0.14 × 119
Mass of KBr = 16.66 grams
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Can anyone help me with this??
In the region labelled as H, weather fronts are most likely to be located. A weather front comprises a line dividing air masses with various differences in their attributes,
A weather front comprises a line dividing air masses with various differences in their attributes, including air density, winds, temperatures, or humidity. These variations frequently result in disturbed or unstable weather at the boundary.
For instance, although warm fronts usually arrive accompanied by stratiform precipitation with fog, cold fronts can bring belts of thunderstorms with cumulonimbus precipitation and be preceded by squall lines. Dry lines, which are softer humidity gradients, can cause severe weather in the summer. In the region labelled as H, weather fronts are most likely to be located.
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Thank you in advance.
The rate expression is k [BF₃]²[NH₃], overall order is 3, rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.
How to calculate rate expression?The rate expression for this reaction can be written as:
Rate = k [BF₃]^m[NH₃]^n
The method of initial rates can be used to calculate the values of m and n. When we compare experiments 1 and 2, notice that halving the concentration of [NH3] reduces the initial rate of reaction. This suggests that the reaction is first order in terms of [NH3], implying that n = 1.
When comparing experiments 1 and 3, notice that increasing the concentration of [BF3] by a factor of 2.5 increases the initial rate of reaction by a factor of 9.39 (i.e., 2.13/0.227). This shows that the reaction is about second order in relation to [BF3], i.e., m 2.
Therefore, the rate expression for the reaction is:
Rate = k [BF₃]²[NH₃]
The overall order of the reaction is m + n = 3.
Using the data from experiment 4, substitute the values and solve for k:
Rate = k [BF₃]²[NH₃]
1.02 × 10-¹ = k (3.00 × 10⁻¹)² (1.00 x 10⁻¹)
k = 1.02 × 10-¹ / (3.00 × 10⁻¹)² (1.00 x 10⁻¹)
k = 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹
Therefore, the rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.
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For each of the following, convert the word equation into a formula equation AND balance it using the tally method!
*The answers have been provided for you, please upload work showing your use of the tally method to earn credit!*
aluminum + nickel(II) sulfate --> aluminum sulfate + nickel
2Al + 3Ni(SO4) --> Al2(SO4)3 + 3Ni
ammonium phosphate + magnesium nitrite --> magnesium phosphate +ammonium nitrite
2(NH4)3(PO4) + 3Mg(NO2)2 --> Mg3(PO4)2 + 6(NH4)(NO2)
lead + nitrogen --> lead(IV) nitride
3Pb + 2N2 --> Pb3N4
Answer:
For aluminum + nickel(II) sulfate --> aluminum sulfate + nickel:
Al + NiSO4 --> Al2(SO4)3 + Ni
Using the tally method:
Al: 2
Ni: 1
S: 3
O: 12
The balanced equation is:
2Al + 3Ni(SO4) --> Al2(SO4)3 + 3Ni
For ammonium phosphate + magnesium nitrite --> magnesium phosphate + ammonium nitrite:
(NH4)3PO4 + Mg(NO2)2 --> Mg3(PO4)2 + 6NH4NO2
Using the tally method:
N: 6
H: 28
P: 2
O: 16
Mg: 3
The balanced equation is:
2(NH4)3(PO4) + 3Mg(NO2)2 --> Mg3(PO4)2 + 6(NH4)(NO2)
For lead + nitrogen --> lead(IV) nitride:
3Pb + 2N2 --> Pb3N4
Using the tally method:
Pb: 3
N: 8
The balanced equation is:
3Pb + 2N2 --> Pb3N4
Explanation:
What is the pH of 6.00 M H2CO3 if it has 7% dissociation? SHOW YOUR WORK!!!
From the calculation, the pH of the solution can be obtained as 0.39.
What is the pH of the solution?We first have to obtain the dissociation constant of the solution;
α = √Ka/C
Ka = Cα^2
Ka = 4.9 * 10^-3 * 6
Ka = 0.0294
Then we have that;
0.0294 = [x]^2/[6 - x]
0.0294 * [6 - x] = x^2
0.1764 - 0.0294 x = x^2
x^2 + 0.0294 x - 0.1764 =0
x = 0.41 M
Thus the pH of the solution is;
pH = -log[0.41]
= 0.39
pH is an important parameter in chemistry, biology, and many other fields, as it can affect the behavior and properties of substances and reactions.
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What volume will 5.00 mol of an ideal gas occupy at 25.0 C. and 153 kPa of pressure?
Answer:
5.00 mol of an ideal gas will occupy 103.6 L at 25.0 C and 153 kPa of pressure.
Explanation:
Using the ideal gas law, PV=nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin, we can solve for V.
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15 K. Therefore, the temperature is 25.0 + 273.15 = 298.15 K.
Next, we can plug in the values we know:
PV = nRT
(153 kPa) V = (5.00 mol) (8.31 J/mol*K) (298.15 K)
Simplifying:
V = (5.00 mol) (8.31 J/mol*K) (298.15 K) / (153 kPa)
V = 103.6 L
Therefore, 5.00 mol of an ideal gas will occupy 103.6 L at 25.0 C and 153 kPa of pressure.
2. Label the different layers of the atmosphere and the separating boundaries between each layer.
The layers of the atmosphere with the boundaries that are between them are shown in the image attached here.
Layers of the atmosphereThe troposphere, which rises up to an altitude of roughly 10-15 km, is the layer that is closest to the Earth's surface. The majority of the Earth's air mass is contained here, which is also where weather happens.
The tropopause marks the transition from the troposphere to the stratosphere, the next layer. It designates the altitude at which the temperature stops dropping.
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Adding 1.64 g of naphthalene to 80.00 g of cyclohexane, C.H₁2() lowers the freezing point from
5.53°C to 1.37°C. If K, for cyclohexane is 19.9°C/m, calculate the molar mass of the naphthalene.
A cycloalkane with the chemical formula C6H12 is cyclohexane. Non-polar describes cyclohexane.
Thus, Colorless and combustible, cyclohexane has a characteristic detergent-like stench that is reminiscent of cleaning products.
The industrial manufacturing of adipic acid and caprolactam, which are building blocks for nylon, uses cyclohexane primarily.
cyclohexanone and cyclohexanol that serves as the starting point for the production of adipic acid and caprolactam, which are both precursors to nylon. Annual production of cyclohexanone and cyclohexanol totals several million kilos.
Thus, A cycloalkane with the chemical formula C6H12 is cyclohexane. Non-polar describes cyclohexane.
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Step 1: A↽−−⇀B+C equilibrium Step 2: C+D⟶E slow Overall: A+D⟶B+E
Determine the rate law for the overall reaction, where the overall rate constant is represented as
The rate law for the reaction can be written as -
rate= k[C] [D]
The rate of reaction or reaction rate is the speed at which reactants are converted into products.
When we talk about chemical reactions, it is a given fact that rate at which they occur varies by a great deal. Some chemical reactions are nearly instantaneous, while others usually take some time to reach the final equilibrium.
The rate of the reaction is governed by the slow step.
This gives the rate equation-
rate= k[C] [D]
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The products of a gamma decay reaction
are shown below. What was the parent
isotope of this nuclear reaction?
22 0
Ne + Y
10 0
The product of the gamma decay reaction is 22/10 Ne
What is gamma decay ?Gamma radiation, commonly referred to as gamma decay, is a type of radioactive decay in which an unstable atomic nucleus generates gamma rays, a form of extremely high-energy electromagnetic radiation.
Gamma rays are neutral and have no mass, in contrast to alpha and beta particles, which have mass and charge.
The result of the gamma decay would not alter the parent nuclei thus the product would have the same mass and charge as the parent nucleus.
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determine whether each molecule will have hydrogen bonding. defend your answer with evidence from your notes.
A - Has no hydrogen bond since hydrogen is not bonded to an electronegative element
B - Has a hydrogen bond due to the N- H bond
C - Has a hydrogen bond due to the O - H bond
D - Has a hydrogen bond due to the H -I bond
What is the hydrogen bond?In a hydrogen bond, a weak electrostatic connection is created when the positively charged hydrogen atom is drawn to the negatively charged electronegative atom.
This contact is stronger than van der Waals forces but not as strong as covalent or ionic bonds or other intermolecular forces.
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When 1500J of energy is lost from a 120 gram object, the temperature decreased from 45° C to 40° C. What is the specific heat of the object?
Specific Heat = 2.5 J/g•°C
We can set up an equation to solve this.
(Energy) = (mass)•(specific heat)•(change in temperature)
Plug in what you know
-1500J = 120g•(c)•(40°C-45°C)
[The energy is negative because the energy is lost, and the change in temperature can be found by subtracting the initial temperature from the final temperature (T(Final) - T(Initial))]
Solve
-1500J = 120g•(c)•(40°C-45°C)
-1500J = 120g•(c)•(-5°C)
-1500J = (-600g•°C)•(c)
(-1500J) / (-600g•°C) = c
c = 2.5 J/g•°C
Readiness Test: Populations, Communities, and Ecosystems
There are unique interactions and relationships which are involved in the transportation of energy. The energy which is produced and captured once is distributed throughout various living organisms. This transfer of energy is called the food web.
The decomposers of the food web are known as the organisms which tend to degrade the dead living organisms of an ecosystem. Some examples of decomposers are fungus, bacterium or invertebrates, etc.
The decomposers help in the energy flow of the ecosystem and help in replenishing the fertility of soil through secreting the digestive enzymes over dead plants and animals and provide nutrients to the soil.
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Hydrogen sulfide will burn \in three different ways, depending upon the amount of oxygen present. In one reaction, sulfur dioxide and water are produced. In the second reaction, water, sulfur dioxide, and sulfur are produced. The third reaction produces water and sulfur. What is maximum amount of sulfur dioxide that can be produced with 50.0g of hydrogen
The term mole concept is used here to determine the mass of sulfur dioxide.The mass of sulfur dioxide obtained from 50.0 g of hydrogen sulphide is 93.93 g.
One mole of a substance is defined as that quantity of it which contains as many entities as there are atoms exactly in 12 g of carbon - 12. The formula used to calculate the number of moles is:
Number of moles = Given mass / Molar mass
Here balanced reaction is:
2H₂S + 3O₂ → 2SO₂ + 2H₂O
n (H₂S) = n(SO₂)
n = 50.0 / 34.1 = 1.466
Molar mass of sulfur dioxide = 64.067 g/mol
Mass = n × M
m = 1.466 × 64.067 = 93.93 g
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Why does increasing the temperature increase the rate of a reaction?
When temperature increases, the particles have more energy
Increasing temperature prevents products from colliding
Particles moving with more energy will collide more often
Warm temperatures are more comfortable
Particles are more likely to collide with enough energy to form an activated complex
The correct answer is "Particles are more likely to collide with enough energy to form an activated complex," which is the last option, as when the temperature of a reaction increases, the average kinetic energy of the particles increases.
Chemical reactions require a certain amount of energy to occur. This energy is called activation energy. In order for reactant molecules to undergo a chemical reaction, they must overcome the activation energy barrier. This barrier is like a hill that reactants must climb over to reach the product state. One way to provide the energy required to overcome the activation energy barrier is to increase the temperature of the system. This is because temperature is a measure of the average kinetic energy of particles in a system. As the temperature increases, the kinetic energy of the particles also increases. This means that they are moving faster and colliding with greater energy.
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Change the following word equations into formula equations AND balance them using the tally method.
*The answers have been provided under each equation, all you have to do is show your work using the tally method!*
a. copper(I) iodate + potassium carbonate → potassium iodate + copper(I)carbonate
2Cu(IO3) + K2(CO3) → 2K(IO3) + Cu2(CO3)
b. iron(III) carbonate + cesium → cesium carbonate + iron
Fe2(CO3)3 + 6Cs → 3Cs2(CO3) + 2Fe
c. sodium chlorate → sodium chloride + oxygen
2Na(ClO3) → 2NaCl + 3O2
The balanced chemical equations with suitable coefficients are 2 Cu(IO₃) + K₂(CO₃) → 2 KIO₃ + Cu₂CO₃, Fe₂(CO₃)₃ + 6 Cs → 3 Cs₂(CO₃) + 2 Fe,
2 NaClO₃ → 2 NaCl + 3 O₂.
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 .
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Help with this
A scientist discovers a new object in the Kuiper Belt and claims it is a dwarf planet. What would have to be true about the object for the scientist to be correct?
A The object would have to be irregularly shaped, have an atmosphere, and be traveling in a path with other objects.
B The object would have to be irregularly shaped, orbit the Sun, and be traveling in a path free from other objects.
C The object would have to be spherical, orbit the Sun, and be traveling in a path with other objects.
D The object would have to be spherical, have an atmosphere, and be traveling in a path free from other objects.
Answer:
C The object would have to be spherical, orbit the Sun, and be traveling in a path with other objects.Explanation:
The sphere is a shape which allows for the most amount of volume with the least amount of surface. It is a very efficient way of enclosing objects and also allows for any point on its surface[tex]\large \boxed{\mathrm{Correct \ answer \ Is \ C.}}[/tex]
[tex]\large \boxed{\mathrm{Sorry \ have \ no \ time \ for \ explanation \ so \ sorry!}}[/tex]
Question 4 of 10
How would one make a 2 M solution of a compound?
A. By dissolving 2 g of the compound in 1 L of water
B. By dissolving 2 moles of the compound in 1 L of water
OC. By dissolving 1 molar mass of the compound in 2 L of water
D. By dissolving 1 mole of the compound in 2 L of water
SUBMIT
2 M solution of a compound can be made by dissolving 2g of compounds in 1 liter of water. Either by diluting the appropriate volume of a more concentrated solution also known as a stock solution to the desired final volume or by dissolving a known mass of solute in a solvent, solutions of known concentration can be prepared.
For example, when we add one part by weight of powder to 19 parts by weight of the solvent is used to make a 5% solution. For 1% solution, the addition of 1gm of NaCl to 100L of water. which is the final volume of the solution.
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Luke needs to measure the iron (Fe) content of a 2.470 g steel sample. He starts by completely converting the iron in the steel to Fe2+ and dissolving it in 50.00 mL of water. Then, Luke determines the amount of Fe2+ by reacting it with a standardized aqueous 0.110 M potassium permanganate (KMnO4) solution.
MnO−4+8H++5Fe2+⟶Mn2++5Fe3++4H2O
The titration of the Fe2+ solution requires 39.23 mL KMnO4 solution to reach the end point. What is the percentage by weight of iron (Fe) in the steel sample?
The balanced equation for the reaction between [tex]Fe^+^2[/tex] and KMnO₄ is: MnO₄ + 8H+ + 5 [tex]Fe^+^2[/tex]⟶ [tex]Mn^+^2[/tex] + 5[tex]Fe^3^+[/tex] + 4H₂O. Therefore, the percentage by weight of iron in the steel sample is 1.94%.
Molarity of KMnO₄ = 0.110 M, Volume of KMnO₄ used = 39.23 mL = 0.03923 L
moles of [tex]Fe^+^2[/tex] = 0.110 M x 0.03923 L / 5 = 0.000856 moles
The mass of iron in the steel sample can be calculated from the number of moles of [tex]Fe^+^2[/tex]:
moles of Fe = 0.000856 moles [tex]Fe^+^2[/tex] x 1 mole Fe / 1 mole [tex]Fe^+^2[/tex]
= 0.000856 moles Fe
The mass of Fe can be calculated from the number of moles of Fe:
mass of Fe = 0.000856 moles Fe x 55.85 g/mol = 0.0479 g Fe
Finally, the percentage by weight of iron in the steel sample can be calculated as:
% Fe = (mass of Fe / mass of steel sample) x 100%
mass of steel sample = 2.470 g
% Fe = (0.0479 g / 2.470 g) x 100% = 1.94%
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What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?
Select one:
A. 13.22 g
O B. 25.0 g/mol
O C. 32.0 g
O D. 33.22 g/mol
O E. 39.0 g
The principal components of gasoline is 702 g of octane.
Thus, Crude oil and other petroleum liquids are used to create gasoline, a fuel. The majority of gasoline is utilized in car engines.
For retail sale at gas stations, finished motor gasoline is produced at petroleum refineries and blending facilities.
In order to create finished motor gasoline, petroleum refineries primarily produce gasoline blendstocks, which must be blended with other liquids. In order to create finished motor gasoline in various grades and formulas, gasoline blendstocks, finished gasoline, and fuel ethanol are combined at blending terminals.
Thus, The principal components of gasoline is 702 g of octane.
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If a sulfuric acid (H₂SO₄, MM = 98.09 g/mol) solution is completely neutralized with 61.37 mL of the NaOH solution above (from question 2) what was the mass in grams of sulfuric acid in the solution. Remember, you need a balanced chemical reaction for stoichiometry. (do not forget about SF)
Taking into account the reaction stoichiometry, 3.00713 grams of H₂SO₄ is completely neutralized with 61.37 mL of the NaOH.
Reaction stoichiometryIn first place, the balanced reaction is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
H₂SO₄: 1 moleNaOH: 2 moles Na₂SO₄: 1 moleH₂O: 2 molesThe molar mass of the compounds is:
H₂SO₄: 98 g/moleNaOH: 40 g/moleNa₂SO₄: 142 g/moleH₂O: 18 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
H₂SO₄: 1 mole× 98 g/mole= 98 gramsNaOH: 2 moles× 40 g/mole= 80 gramsNa₂SO₄: 1 mole× 142 g/mole= 142 gramsH₂O: 2 moles× 18 g/mole= 36 gramsMass of H₂SO₄ requiredA sulfuric acid solution is completely neutralized with 61.37 mL (0.06137 L) of the NaOH solution. Assuming the concentration of the NaOH solution is 1.0 M, the moles of NaOH that react is calculated as:
Moles of NaOH reacting = concentration× volume in liters
Moles of NaOH reacting = 1 M× 0.06137 L
Moles of NaOH reacting = 0.06137 moles
Now, the following rule of three can be applied: If by reaction stoichiometry 2 moles of NaOH react with 98 grams of H₂SO₄, 0.06137 moles of NaOH react with how much mass of H₂SO₄?
mass of H₂SO₄= (0.06137 moles of NaOH× 98 grams of H₂SO₄)÷2 moles of NaOH
mass of H₂SO₄= 3.00713 grams
Finally, 3.00713 grams of H₂SO₄ is required.
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2.50 g of As2O3 are titrated with 38.5 mL of KMnO4 to reach the end point.
5As2O3(s)+4MnO−4(aq)+9H2O(l)+12H+(aq)⟶10H3AsO4(aq)+4Mn2+(aq)
Calculate the concentration of the KMnO4 solution.
50 g of As[tex]_2[/tex]O[tex]_3[/tex] are titrated with 38.5 mL of KMnO[tex]_4[/tex] to reach the end point. 0.26M is the concentration of the KMnO[tex]_4[/tex] solution.
Concentration in chemistry refers to the quantity of a material in a certain area. The ratio of the solute within a solution to the solvent or whole solution is another way to define concentration. In order to express concentration, mass in unit volume is typically used.
The solute concentration can, however, alternatively be stated in moles or volumetric units. Concentration may be expressed as per unit mass rather than volume.
5As[tex]_2[/tex]O[tex]_3[/tex](s)+4MnO[tex]_4[/tex]⁻(aq)+9H[tex]_2[/tex]O(l)+12H⁺(aq)⟶10H[tex]_3[/tex]AsO[tex]_4[/tex](aq)+4Mn[tex]_2[/tex]⁺(aq)
the stoichiometry ratio between As[tex]_2[/tex]O[tex]_3[/tex] and MnO[tex]_4[/tex]⁻ is 5:4
0.0126 moles of As[tex]_2[/tex]O[tex]_3[/tex] will react with 4/5×0.0126 moles = 0.01008moles
0.01008moles of MnO[tex]_4[/tex]⁻ is present in 38mL
concentration of KMnO[tex]_4[/tex]= moles×volume
= 0.010/38×1000
=0.26M
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Determine E° for a galvanic (voltaic) cell if ∆G° = -6.3 kJ/mol and n = 3. (F = 96,500 J/(V･mol))
The E° for a galvanic cell is 0.000217 volts if ∆G° = -6.3 kJ/mol and n = 3. (F = 96,500 J/(V･mol).
A galvanic cell or voltaic cell, named after the scientists Luigi Galvani and Alessandro Volta, respectively, is an electrochemical cell in which an electric current is generated from spontaneous Oxidation-Reduction reactions. A common apparatus generally consists of two different metals, each immersed in separate beakers containing their respective metal ions in solution that are connected by a salt bridge or separated by a porous membrane.
E°=ΔG°/-nF= -6.3/-3×96500=0.000217 V.
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 If you mixed three plastic cup's contents containing 0,05 L of 1M Kool-Aid, 0.05 L of 2.5M Kool-Aid, and 0.05 L of 0.5M Kool-Aid, what would the molarity be of the resulting solution?
M = 1.183
Molarity is moles/Liter, so setting up some equations gives you
1) 1M = Xmol/0.05L
2) 2.5M = Xmol/0.05L
3) 0.5M = Xmol/0.05L
Solve for each X
1) X = 0.05mol
2) X = 0.125mol
3) X = 0.0025mol
Now add all the moles and Liters (separately)
XM = (0.05+0.125+0.0025)mol/(0.05+0.05+0.05)L
XM = 0.1775mol/0.15L
X = 1.183M
What is the difference in energy levels of the sodium atom if emitted light has a wavelength of 589 nm?
Calculating equilibrium concentrations when the net reaction proceeds forward
Consider mixture B, which will cause the net reaction to proceed forward.
Concentration (M)initial:change:equilibrium:[XY]0.500−x0.500−xnet→⇌[X]0.100+x0.100+x+[Y]0.100+x0.100+x
The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced.
Based on a Kc value of 0.140 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?
We must utilize the reaction's equilibrium equation to determine the equilibrium concentrations:
Kc = [X][Y]/[XY]
replacing the specified values:
0.140 = (0.100+x)(0.100+x)/(0.500-x)
Expanding and condensing:
0.14 = (0.01 + 0.2x + x^2)/(0.5 - x)
0.07 - 0.14x = x^2 + 0.2x + 0.01
Changing the order and applying the quadratic formula to find x
x^2 + 0.34x - 0.06 = 0
x = 0.193 or -0.533
X cannot be negative because it denotes a change in concentration. Consequently, x = 0.193 M.
It is now possible to determine the equilibrium concentrations:
[XY] = 0.500 - x = 0.307 M
[X] = [Y] = 0.100 + x = 0.293 M
Therefore, the equilibrium concentrations of XY, X, and Y are 0.307 M, 0.293 M, and 0.293 M, respectively.
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Q1-50.0-mL sample of 0.200 M sodium hydroxide is titrated with
0.200 M Hydrochloric acid acid. Calculate:
(a) the pH after adding 30.00 mL of HCl.
(b) the pH at the equivalence point
(c) the pH after you add a total of 60.0 mL of 0.200 M HCl
Round pH values to two decimal places.
Q2- A 40.0-mL sample of 0.100 M HNO2 is titrated with 0.200 M
KOH. Calculate:
(a) the volume required to reach the equivalence point
(b) the pH after adding 5.00 mL of KOH
(c) the pH at one-half the equivalence point
(d) the pH at the equivalence point for the titration between
HNO2 and KOH
The acidity or alkalinity of a solution depends upon its hydronium ion concentration and hydroxide ion concentration. The pH scale is introduced by the scientist Sorensen.
The pH of a solution is defined as the negative logarithm to the base 10 of the hydronium ion concentration in moles per litre.
a. Concentration of NaOH = 0.200 M
Concentration of HCl = 0.200 M
Total volume = 50 + 30 = 80 mL = 0.08 L
The neutralization reaction between NaOH and HCl is:
OH⁻ + H⁺ ----------> H₂O
Then,
n(H⁺) = 0.03 × 0.200 = 0.006 moles
n(OH⁻) = 0.05 × 0.200 = 0.01 moles
Remaining moles of OH⁻ = 0.01 - 0.006 = 0.004
pOH = -log [OH⁻] = 2.39
pH + pOH = 14
pH = 14 - pOH = 14 - 2.39 = 11.61
b. pH at the equivalence point is about pH 7.
c. Total volume = 0.11 L
n(H⁺) = 0.06 × 0.200 = 0.012 moles
n(OH⁻) = 0.05 × 0.200 = 0.01 moles
Remaining moles of H⁺ = 0.012 -0.01 = 0.002
Concentration of H⁺ is 0.002 M / 0.11 L = 0.01818 moles/L
pH = - log [ H⁺ ]
pH = - log [ 0.01818 ]
pH = 1.74
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5. Which of the following are correct?
I. Electrons are attracted towards higher potential.
II. Protons are attracted towards higher potential.
III. Electrons are attracted towards lower potential.
IV. Protons are attracted towards lower potential.
(a) I & II
(b) I & III
(c) II & III
(d) I & IV
The statements that are correct is that electrons are attracted towards higher potential and protons are attracted towards lower potential. So the correct option is d. I & IV
What are electrons and protons?Electrons are particles that are negatively charged so if it is placed in an electric field it will move from negative to positive, due to its negative charge, then it will attract a higher potential.
Protons are particles that are positively charged so if it is placed in an electric field it will move from a positive to a negative side, then due to its positive charge, it will be attracted to a lower potential.
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Part D
Now predict the shape of a graph that shows velocity in the x direction (v) versus time
The shape of a graph that shows the velocity in the x direction (v) versus time when the object is accelerating in the x direction would be a curve with a non-zero slope.
What is the velocity of a body?The rate at which an object's displacement changes in relation to a frame of reference is its velocity. The direction of the movement of the body or the object is defined by its velocity.
According to the equation v = s/t, velocity (v) is a vector quantity that quantifies displacement (or change in position, s), over the change in time (t).
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Answer: I expect a horizontal line (slope 0). As time increases, the velocity in the x direction stays about the same.
Explanation: Edmentum sample answer
A. What volume of base was needed to reach the equivalence point. B. What is the pH at the equivalent point?
From the titration curve that have been shown in the image, the equivalence point is 50 mL
What is the equivalence point on a titration curve?At the equivalence point on a titration curve, the amount of titrant added is chemically equivalent to the amount of analyte in the sample being evaluated. As a result of the reaction between the titrant and analyte at this point, the entire analyte has been neutralized by the titrant.
You can locate the equivalence point by plotting the pH or any relevant aspect of the sample under examination as a function of the volume of titrant used.
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Completely describe the electrolytic cell corresponding to the following equation. (Hint: you may need to combine 2 half reactions from Table 17-1 to make one of the half reactions for this cell)
Cr2O7^2– + I^– → Cr^3+ + IO3^–
With work please
The first half-reaction is the oxidation of Cr2O7^2– to Cr^3+ and the second half-reaction is the reduction of I^– to IO3^–. When combined, the overall reaction is Cr2O7^2– + I^– → Cr^3+ + IO3^–.
The electrolytic cell consists of two electrodes, one anode and one cathode, both of which are immersed in an electrolyte solution. At the anode, the Cr2O7^2– ions are oxidized to Cr^3+ ions, releasing electrons into the external circuit.
At the cathode, the I^– ions are reduced to IO3^– ions, and the electrons from the external circuit are used to drive the reaction. The electrolyte solution must contain both Cr2O7^2– and I^– ions in order to facilitate the transfer of electrons between the electrodes.
The overall reaction is driven by the potential difference between the anode and the cathode, which is created by the flow of electrons through the external circuit.
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