If you have 3kg of lead and put 3840 joules of heat energy into that mass of lead. Taking
careful measurements, you observe the temperature of the lead rise from 40C to 50C.
What is the specific heat capacity of lead?
Answer : The specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]
Explanation :
Formula used:
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = 3840 J
m = mass of lead = 3 kg = 3000 g
c = specific heat capacity of lead = ?
[tex]T_1[/tex] = initial temperature = [tex]40^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]50^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]3840J=3000g\times c\times (50-40)^oC[/tex]
[tex]c=0.128J/g^oC[/tex]
Therefore, the specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]
Given: N2H4(l) + O2(g) LaTeX: \longrightarrow⟶ N2(g) + 2H2O(g) ΔH°1 = –543 kJ·mol–1 2H2(g) + O2(g) LaTeX: \longrightarrow⟶ 2H2O(g) ΔH°2 = –484 kJ·mol–1 N2(g) + 3H2(g) LaTeX: \longrightarrow⟶ 2NH3(g) ΔH°3 = –92 kJ·mol–1 What is the standard enthalpy change for the following reaction? 2NH3(g) LaTeX: \longrightarrow⟶ N2H4(l) + H2(g) Group of answer choices
Answer:
Explanation:
Given reaction
N₂H₄ + O₂ ⇒ N₂ + 2H₂O ΔH₁ = -543 KJ ---------- ( 1 )
2H₂ + O₂ ⇒ 2H₂O ΔH₂ = -484 KJ ---------- ( 2 )
N₂ + 3 H₂ ⇒ 2NH₃ ΔH₃ = -92 KJ -----------( 3 )
( 1 ) - ( 2 ) +( 3 )
N₂H₄ + O₂ - 2H₂ - O₂ +N₂ + 3 H₂ ⇒ N₂ + 2H₂O - 2H₂O +2NH₃
ΔH = -543 + 484 -92 = -151 KJ
N₂H₄ + H₂ ⇒ 2NH₃ ΔH = -151 KJ .
2NH₃ ⇒ N₂H₄ + H₂ ΔH = + 151 KJ
Part 1: Write the complete balanced MOLECULAR equation (including all states of matter) for the precipitation reaction that occurs between aqueous copper(II) chloride and aqueous sodium hydroxide.
Part 2: Write the NET IONIC EQUATION for the REVERSE reaction in Part 1 (include all states of matter).
Part 3: Write the equilibrium expression for the net ionic equation written in Part 2.
Answer:
For 1: The molecular equation is [tex]CuI_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaI(aq)[/tex]
For 2: The net ionic equation is [tex]Cu(OH)_2(s)\rightarrow Cu^{2+}(aq)+2I^{-}(aq)[/tex]
For 3: The equilibrium constant expression is [tex]K_{eq}=[Cu^{2+}][I^-]^2[/tex]
Explanation:
For 1:A molecular equation is defined as the chemical equation in which the ionic compounds are written as molecules rather than component ions.
The molecular equation for the reaction of copper (II) iodide and sodium hydroxide is given as:
[tex]CuI_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaI(aq)[/tex]
For 2:Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of copper hydroxide and sodium iodide is given as:
[tex]2NaI(aq)+Cu(OH)_2(s)\rightarrow 2NaOH(aq)+CuI_2(aq)[/tex]
Ionic form of the above equation follows:
[tex]2Na^{+}(aq)+2I^{-}(aq)+Cu(OH)_2(s)\rightarrow 2Na^+(aq)+2OH^-(aq)+Cu^{2+}(aq.)+2I^{-}(aq.)[/tex]
As, sodium and iodide ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
[tex]Cu(OH)_2(s)\rightarrow Cu^{2+}(aq)+2I^{-}(aq)[/tex]
For 3:The expression of equilibrium constant for the net ionic equation above follows:
[tex]K_{eq}=[Cu^{2+}][I^-]^2[/tex]
Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.
Construct a simulated 1H NMR spectrum, including proton integrations, for CH3OC(CH2OCH3)3 (see Hint). Drag the appropriate splitting patterns to the approximate chemical shift positions; place the integration values in the small bins above the associated chemical shift. Splitting patterns and integrations may be used more than once, or not at all, as needed. Likewise, some bins might remain blank. Note that peak heights are arbitrary and do not indicate proton integrations.
Answer:
The drawing of the structure is found in diagram 1 of the attached figure.
Explanation:
Diagram 1 shows that three different types of protons are found in the structure. The nine hydrogen atoms have a similar behavior, the six hydrogen atoms also have a similar behavior and finally, the three hydrogen atoms adjacent to oxygen have a similar behavior. The number of peaks are as follows:
9H = singlet peak = between 3 and 4 ppm
6H = singlet peak = 4 ppm
3H = singlet peak = 3 ppm.
The 9 protons are around 3.5 ppm and the 6 hydrogen atoms show a peak at 4 ppm, and finally, the 3 protons have a peak around 3 ppm. Therefore, the corresponding drawing can be seen in diagram 2.
To construct a simulated 1H NMR spectrum for CH3OC(CH2OCH3)3, analyze the structure and determine the chemical shifts and integration values for each proton.
Explanation:In order to construct a simulated 1H NMR spectrum for CH3OC(CH2OCH3)3, we need to analyze the structure of the molecule and determine the chemical shifts and integration values for each proton. Let's break down the molecule:
The CH3 group attached to the oxygen atom will appear as a singlet at around 3.2 ppm with an integration value of 3, as it represents three equivalent protons.The CH2 group attached to the oxygen atom will appear as a quartet at around 3.6 ppm with an integration value of 2, as it represents two equivalent protons that are adjacent to a CH3 group.The CH2 group attached to the CH2 group will appear as a septet at around 3.8 ppm with an integration value of 2, as it represents two equivalent protons that are adjacent to two CH3 groups.The CH2 group attached to another CH2 group will also appear as a septet at around 3.8 ppm with an integration value of 2, as it represents two equivalent protons that are adjacent to two CH3 groups.The CH3 groups attached to the CH2 groups will appear as singlets at around 4.0 ppm with an integration value of 9, as each CH3 group represents three equivalent protons.By analyzing the structure and applying the appropriate splitting patterns and chemical shifts, we can construct a simulated 1H NMR spectrum for CH3OC(CH2OCH3)3.
Learn more about 1H NMR spectrum here:https://brainly.com/question/30667404
#SPJ11
Cathodic protection of a metal pipe against corrosion usually entails __________.
Options:
A. attaching an active metal to make the pipe the anode in an electrochemical cell.
B. coating the pipe with another metal whose standard reduction potential is less negative than that of the pipe.
C. attaching an active metal to make the pipe the cathode in an electrochemical cell.
D. attaching a dry cell to reduce any metal ions which might be formed.
E. coating the pipe with a fluoropolymer to act as a source of fluoride ion (since the latter is so hard to oxidize).
Answer:
C. attaching an active metal to make the pipe the cathode in an electrochemical cell.
Explanation:
Cathodic protection is a technique which helps in controlling the increased rate of corrosion of a metal surface by making it the cathode of an electrochemical cell. It connects the metal to be protected to a more easily corroded which is usually referred to as the sacrificial metal to act as the anode.
This technique preserves the metal by providing a highly active metal that can act as an anode and provide free electrons. By introducing these free electrons, the active metal sacrifices its ions and keeps the less active steel from corroding.
Water standing in the open at 33.0°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (557 cal/g) is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per gram. (a) Find ε. (b) What is the ratio of ε to the average kinetic energy of H2O molecules, assuming the latter is related to temperature in the same way as it is for gases?
Answer:
a) ε = 6.961 × 10⁻²⁰Joules
b)The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642
Explanation:
a) The formula to be used is given below as :
Heat of Vapourisation(Lv) = εn
Where: ε = is the average energy of the escaping molecules and
n = is the number of molecules per gram
The first step is to convert 557 cal/g to joules/kilogram(j/kg)
1 cal/g = 4186.8j/kg
557cal/g = ?
We cross multiply
557cal/g × 4186.8j/kg
= 2332047.6j/kg
Therefore, 557 cal/g = 2332047.6j/kg
ε = Lv/n
ε = LvM/n
Where Lv = 2332047.6j/kg
M = 0.018kg/mol
n = 6.03 × 10²³mol
ε =( 2332047.6j/kg × 0.018kg/mol) ÷ 6.03 × 10²³mol
= 6.961336119 × 10⁻²⁰Joules
Approximately, ε = 6.961 × 10⁻²⁰Joules
b) Kinetic energy = (3/2)KT
The ratio of ε to Kinetic energy = ε/(3/2)kT = 2ε /3kT
Where ε = 6.961 × 10⁻²⁰Joules
k = 1.38× 10⁻²³ Joules/kelvin
T = 33°C , which will be converted to kelvin as
33°C + 273K
= 306K
The ratio of ε to Kinetic energy will be calculated as
2ε /3kT
= (2×6.961 × 10⁻²⁰ Joules) ÷ (3 × 1.38× 10⁻²³Joules/kelvin × 306K)
= 10.642
Hence , The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642
One mole of an atom of a substance is equal to
Answer:
One mole of a substance is equal to 6.022 × 10²³ units of that substance (atoms, molecules, or ions)
Explanation:
This number is Avogadro's number. The concept of the mole can be used to convert between mass and number of particles. its used to compare very large numbers.
100!!!POINTS PLZ HELP Explain (on the molecular level) what pumping a tire with air will do to
the pressure
Answer:
Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.
The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure 2. Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them.
Amino acids that are ketogenic lead to ketone bodies that do not contribute to diabetic ketoacidosis. that are glucogenic all produce oxaloacetate and then glucose. are either ketogenic or glucogenic, but never both. are solely ketogenic in only three cases. g
Answer:
that are glucogenic all produce oxaloacetate and then glucose.
Explanation:
The major aim of protein catabolism during a state of starvation is to provide the glucogenic amino acids especially alanine and glutamine, that serve as substrates for endogenous glucose production ie gluconeogenesis in the liver.
Glucogenic amino acid are known for their production of oxaloacetate and then glucose.
Consider the following reaction at equilibrium. What will happen if SO2SO2 is added to the reaction? 4FeS2FeS2(s) + 11O2O2(g) ⇌⇌ 2Fe2O3Fe2O3(s) + 8SO2SO2(g) Consider the following reaction at equilibrium. What will happen if is added to the reaction? 4(s) + 11(g) 2(s) + 8(g) The equilibrium constant will increase. The equilibrium will change in the direction of the reactants. The equilibrium will change in the direction of the products. No change in equilibrium is observed. The equilibrium constant will decrease.
Answer:
The equilibrium will change in the direction of the reactants
Explanation:
The isomerization of methylisonitrile to acetonitrile (CH3NC(g) ???? CH3CN) is first order in CH3NC. The rate constant for the reaction is 9.45 x 10-5 s-1 at 478 K. What is the half-life of the reaction when the initial concentration of CH3NC is 0.0300 M?
Answer:
The half life is [tex]H_{1/2}= 7333.3sec[/tex]
Explanation:
The half life of a first order reaction is mathematically represented as
[tex]H_{1/2} = \frac{0.693}{Rate Constant }[/tex]
Substituting [tex]9.45 * 10^{-5}s^{-1}[/tex] for the rate constant
[tex]H_{1/2} = \frac{0.693}{9.45*10^{-5}}[/tex]
[tex]H_{1/2}= 7333.3sec[/tex]
Final answer:
The half-life of the isomerization of methylisonitrile to acetonitrile, which is a first-order reaction with a rate constant of 9.45 x 10⁻⁵ s⁻¹, is approximately 7330 seconds.
Explanation:
The student's question asks for the calculation of the half-life of the reaction where methylisonitrile isomerizes to acetonitrile (CH₃NC to CH₃CN). This reaction is first order with respect to methylisonitrile, and the rate constant k is given as 9.45 x 10⁻⁵ s⁻¹ at 478 K. To find the half-life ([tex]t^{1/2[/tex]), we can use the first-order half-life equation, [tex]t^{1/2[/tex] = (ln 2) / k. Substituting the given rate constant into this equation
[tex]t^{1/2[/tex] = (ln 2) / (9.45 x 10⁻⁵ s⁻¹)
[tex]t^{1/2[/tex] = (0.693) / (9.45 x 10⁻⁵ s⁻¹)
[tex]t^{1/2[/tex] ≈ 7330 seconds
The half-life of the reaction under the given conditions is approximately 7330 seconds.
Consider the reversible reaction.
PCl5↽−−⇀PCl3+Cl2PCl5↽−−⇀PCl3+Cl2
Are the concentrations of phosphorus pentachloride, PCl5,PCl5, and phosphosphorus trichloride, PCl3,PCl3, constant or changing at equilibrium?
a.The concentration of PCl5PCl5 is constant, and the concentration of PCl3PCl3 is changing at equilibrium.
b.The concentration of PCl5PCl5 is changing, and the concentration of PCl3PCl3 is constant at equilibrium.
c.The concentrations of both PCl5PCl5 and PCl3PCl3 are constant at equilibrium.
d.The concentrations of both PCl5PCl5 and PCl3PCl3 are changing at equilibrium.
which one is correct please explain
In a reversible reaction at equilibrium, including the reaction PCl5↽−−⇀PCl3+Cl2, the concentrations of all components, including PCl5 and PCl3, remain constant because the forward and reverse reactions occur at the same rate.
Explanation:In a reversible reaction at equilibrium the concentrations of all reactants and products remain constant. This is because the rate of the forward reaction equals the rate of the reverse reaction. So, in the context of the provided equation PCl5↽−−⇀PCl3+Cl2, when the system reaches equilibrium, no changes are noticeable, meaning the concentrations of phosphorus pentachloride, PCl5, and phosphorus trichloride, PCl3, as well as chlorine (Cl2), are constant. Therefore, the correct answer is (c) The concentrations of both PCl5 and PCl3 are constant at equilibrium.
Learn more about Chemical Equilibrium here:https://brainly.com/question/3920294
#SPJ12
At equilibrium, the concentrations of both phosphorus pentachloride (PCl5) and phosphorus trichloride (PCl3) are constant due to the condition of dynamic equilibrium. The reaction continues to occur, but the amounts of reactants and products remain unchanged.
Explanation:In the context of the reversible reaction PCl5 ↔ PCl3 + Cl2, at equilibrium, both the concentration of phosphorus pentachloride, PCl5, and the concentration of phosphorus trichloride, PCl3, are constant.
This is described as a condition known as dynamic equilibrium, where the forward and reverse reactions are happening at the same rate, meaning the concentrations of PCl5, PCl3, and Cl2 are not changing, even though the reacting process is still happening. Therefore, the correct answer is (c) The concentrations of both PCl5 and PCl3 are constant at equilibrium.
Learn more about Dynamic Equilibrium here:https://brainly.com/question/26034510
#SPJ2
Select the correct value for the indicated bond angle in each of the compounds.
90°, 180°, 109.5°, 120°, <120°, <109.5°
a) O−O−O angle of O3
b) F−B−F angle of BF3
c) F−O−F angle of OF2
d) Cl−Be−Cl angle of BeCl2
e) F−P−F angle of PF3
f) Cl−Si−Cl angle of SiCl4
The bond angles for ozone (O₃), boron trifluoride (BF₃), oxygen difluoride (OF₂), beryllium dichloride (BeCl₂), phosphorus trifluoride (PF₃), and silicon tetrachloride (SiCl₄) are <120°, 120°, <109.5°, 180°, <109.5°, and 109.5° respectively, based on their molecular geometries.
The correct value for the indicated bond angle in each of the compounds is based on the molecular geometry and electron-pair repulsion theory. Here are the answers:
Ozone (O₃): The O-O-O angle in O₃ is expected to be <120° due to its bent molecular structure.
Boron trifluoride (BF₃): The F-B-F angle in BF3 is 120° because of its trigonal planar molecular geometry.
Oxygen difluoride (OF₂): The F-O-F angle in OF2 is expected to be <109.5° because fluoro substituents cause greater repulsion than hydrogens, narrowing the bond angle from the tetrahedral
Beryllium dichloride (BeCl₂): The Cl-Be-Cl angle in BeCl₂ is 180° due to its linear molecular structure.
Phosphorus trifluoride (PF₃): The F-P-F angle in PF₃ is expected to be <109.5° due to its trigonal pyramidal molecular structure.
Silicon tetrachloride (SiCl₄): The Cl-Si-Cl angle in SiCl₄ is 109.5° since it has a tetrahedral molecular geometry.
Propane (C3H8) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. Include all reaction states. → (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.65 g of propane. L
Answer:
11.6 L will be the number of liters of carbon dioxide measured at STP.
Explanation:
The balanced equation for this combustion reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
where 1 mol of propane reacts to 5 moles of oxygen in order to produce 3 moles of carbon dioxide and 4 moles of water.
We assume the oxygen in excess, so the limiting reagent is the propane. Now, we determine the moles: 7.65 g . 1 mol/ 44 g = 0.174 moles
Ratio is 1:3. 1 mol of propane can produce 3 moles of CO₂
Therefore, 0.174 moles will produce (0.174 . 3) / 1 = 0.521 moles of CO₂
As 1 mol of gas is contained in 22.4L at STP conditions, we propose
22.4L / 1 mol = V₂ / 0.521 mol
22.4 L / 1 mol . 0.521 mol = V₂ → 11.6 L
Answer:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)
11.6 L of CO2 will be produced
Explanation:
Step 1: Data given
Mass of propane = 7.65 grams
Molar mass propane = 44.1 g/mol
Burning = combustion reation = adding O2. The products will be CO2 and H2O
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)
Step 3: Calculate moles propane
Moles propane = 7.65 grams / 44.1 g/mol
Moles propane = 0.173 moles
Step 4: Calculate moles CO2
For 1 mol propane we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 0.173 moles propane we'll have 3*0.173 = 0.519 moles CO2
Step 5: Calculate volume of CO2
1 mol = 22.4 L
0.519 moles = 22.4 L * 0.519 = 11.6 L
11.6 L of CO2 will be produced
Which of the following compounds would be consistent with a compound showing an M with an m/z ratio of 43 in a mass spectrometer? Select all that apply.
a. CH3CH2OH
b. CH3CH2NH2
c. propane
d. propene
e. None of the choices are correct.
Answer:
The option (e) None of the choice are correct.
Explanation:
To find [tex]\frac{m}{z}[/tex] ratio in mass spectrometer,
We start calculating from the first option
(a)
[tex]CH_{3}- CH_{2}- OH[/tex]
We know mass of carbon [tex]m= 12[/tex] and atomic number of carbon [tex]Z = 6[/tex]
After calculating we come up with,
[tex]\frac{m}{z} = 46[/tex]
Therefore option (a) is incorrect.
(b)
[tex]CH_{3} -CH_{2}- NH_{2}[/tex]
After calculating we come up with,
[tex]\frac{m}{z} = 45[/tex]
Therefore option (b) is incorrect.
(c)
[tex]CH_{3}- CH_{2}-CH_{3}[/tex]
[tex]\frac{m}{z} = 44[/tex]
Therefore option (c) is incorrect.
(d)
[tex]CH_{3}- CH=CH_{2}[/tex]
[tex]\frac{m}{z} = 42[/tex]
Therefore option (d) is incorrect.
Therefore, the option (e) is correct for this problem
Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that of hydrofluorocarbons. One source of NO· in the stratosphere is supersonic aircraft whose jet engines convert small amounts of N2 and O2 to NO·. Write the propagation steps for the reaction of O3 with NO·.
Answer:
O3 + NO. = O2. + NO2
2O. +NO2 = NO. + 2O2
Explanation:
NO. = Radical
O. & O2. = Radical
Final answer:
Nitric oxide (NO) catalytically destroys ozone (O3) in the stratosphere via a series of propagation steps, contributing to ozone layer depletion and increasing exposure to harmful UV rays.
Explanation:
Nitric oxide (NO) plays a significant role in the destruction of ozone (O3) through a series of reactions in the stratosphere. The presence of NO contributes to the depletion of the ozone layer, which can increase the risk of harmful UV exposure to humans and the environment. The relevant ozone depletion mechanism proceeds with nitric oxide acting as a catalyst through the following propagation steps:
NO + O3 → NO2 + O2NO2 + O3 → NO3 + O2NO3 + NO2 → N2O5These reactions show how NO continues to regenerate and is not consumed, which is why it acts as a catalyst, increasing the rate of ozone decomposition. The nitric oxide is primarily produced by high-flying aircraft and motor vehicles, raising concerns about its environmental impact.
If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate it, what is the acetic acid concentration in the mixture
When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.
Let's consider the neutralization reaction between sodium hydroxide and acetic acid.
NaOH + CH₃COOH ⇒ CH₃COONa + H₂O
32.40 mL of 0.258 M NaOH react. The reacting moles of NaOH are:
[tex]0.03240 L \times \frac{0.258mol}{L} = 8.36 \times 10^{-3} mol[/tex]
The molar ratio of NaOH to CH₃COOH is 1:1. The moles of CH₃COOH required to react with 8.36 × 10⁻³ moles of NaOH are:
[tex]8.36 \times 10^{-3} mol NaOH \times \frac{1molCH_3COOH}{1molNaOH} = 8.36 \times 10^{-3} mol CH_3COOH[/tex]
8.36 × 10⁻³ moles of CH₃COOH are in 1.00 mL of solution. The concentration of CH₃COOH is:
[tex]M = \frac{8.36 \times 10^{-3} mol}{1.00 \times 10^{-3} L} = 8.36 M[/tex]
When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.
Learn more: https://brainly.com/question/2728613
The concentration of acetic acid in the mixture can be found by titrating with NaOH, using a stoichiometric reaction to equate the moles of NaOH to moles of acetic acid, and then calculating the concentration of the acid.
Explanation:This problem is about using acid-base reactions to determine the concentration of acetic acid in a solution. The reaction between the acetic acid and sodium hydroxide (NaOH) is a stoichiometric reaction where one mole of acetic acid reacts with one mole of NaOH. We use that information to calculate the number of moles of acetic acid.
First, we calculate the number of moles of NaOH used in the titration, which is the concentration of NaOH multiplied by the volume used (in liters): moles NaOH = (0.258 mol/L) * (32.40 mL * 1L/1000 mL) = 0.00835 mol NaOH.
Since the reaction is stoichiometric, the number of moles of NaOH is equivalent to the number of moles of acetic acid.
Finally, the concentration of acetic acid is the number of moles divided by the volume of the sample: concentration = moles / volume = 0.00835 mol / 0.001 L = 8.35 M.
Learn more about Acid-Base Titration here:https://brainly.com/question/40172894
#SPJ11
Tooth enamel is composed of hydroxyapatite, Ca5(PO4)3OH (Ksp = 6.8 10-37). Many water treatment plants add F- ion to drinking water, which reacts with Ca5(PO4)3OH to form the more tooth decay-resistant fluorapatite, Ca5(PO4)3F (Ksp = 1.0 10-60). Fluoridated water has resulted in a dramatic decrease in the number of cavities among children. Calculate the solubility of Ca5(PO4)3F in water.
Answer:
6.1 × 10⁻⁸ M
Explanation:
Let's consider the solution of fluorapatite.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
We can relate the solubility (S) with the solubility product constant (Ksp) using an ICE chart.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
I 0 0 0
C +5 S +3 S +S
E 5 S 3 S S
The Ksp is:
Ksp = [Ca²⁺]⁵ × [PO₄³⁻]³ × [F⁻] = (5 S)⁵ × (3 S)³ × S = 84,375 S⁹
[tex]S = \sqrt[9]{\frac{Ksp}{84,375} } = \sqrt[9]{\frac{1.0 \times 10^{-60} }{84,375} } =6.1 \times 10^{-8} M[/tex]
The solubility of fluorapatite compound Ca₅(PO₄)₃F in water is 6.1 × 10⁻⁸M.
What is Ksp?Ksp is the solubility product constant which tells about the relative solubility of the compound which is in equilibrium with their constitute ions.
Chemical reaction for the solubility of fluorapatite compound is:
Ca₅(PO₄)₃F(s) ⇄ 5Ca²⁺(aq) + 3PO₄³⁻(aq) + F⁻(aq)
Let at equilibrium formed concentration of Ca²⁺, PO₄³⁻ and F⁻ are 5x, 3x and x respectively.
Given value of Ksp for Ca₅(PO₄)₃F = 1.0 × 10⁻⁶⁰
Ksp equation for the given reaction will be represented as:
Ksp = [Ca²⁺]⁵[PO₄³⁻]³[F⁻]
On putting above values in the equation we get,
1.0 × 10⁻⁶⁰ = (5x)⁵. (3x)³. (x)
x = 6.1 × 10⁻⁸M
Hence, solubility of Ca₅(PO₄)₃F is 6.1 × 10⁻⁸M.
To know more about Ksp, visit the below link:
https://brainly.com/question/7185695
When a complex ion forms, ______________ arrange themselves around _________, creating a new ion with a charge equal to the sum of the charges of its components. Select the correct answer below: transition metal ions, the ligand ions, the coordinate complex ligands, the central atom none of the above
Answer:
lignands, the central atom/metal ion
Explanation:
When a complex ion forms, ligands arrange themselves around , the central atom/metal ion creating a new ion.
What is an Ion?This is defined as an atom or molecule which possesses an electric charge such as positive or negative.
Complex ion form as a result of ligands arrangement around central atom/metal ion with a charge equal to the sum of the charges of its components.
Read more about Complex ion here https://brainly.com/question/24262383
#SPJ2
Nitric acid is often manufactured from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming of natural gas, in a two-step process. In the first step, nitrogen and hydrogen react to form ammonia: (g)(g)(g) In the second step, ammonia and oxygen react to form nitric acid and water: (g)(g)(g)(g) Write the net chemical equation for the production of nitric acid from nitrogen, hydrogen and oxygen. Be sure your equation is balanced.
Answer: [tex]N_2(g)+3H_2(g)+4O_2(g)\rightarrow 2HNO_3(g)+2H_2O(g)[/tex]
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced chemical reaction for nitrogen and hydrogen react to form ammonia:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex] (1)
The balanced chemical reaction for ammonia and oxygen react to form nitric acid and water:
[tex]NH_3(g)+2O_2(g)\rightarrow HNO_3(g)+H_2O(g)[/tex] [tex]\times 2[/tex]
[tex]2NH_3(g)+4O_2(g)\rightarrow 2HNO_3(g)+2H_2O(g)[/tex] (2)
The net chemical equation for the production of nitric acid from nitrogen, hydrogen and oxygen by adding 1 and 2
[tex]N_2(g)+3H_2(g)+4O_2(g)\rightarrow 2HNO_3(g)+2H_2O(g)[/tex]
Final answer:
The net balanced chemical equation for the manufacturing of nitric acid from nitrogen, hydrogen, and oxygen is N₂(g) + O₂(g) + 3H₂(g) → 2HNO₃(g) + 2H₂O(l). This reflects a combination of the Haber process for ammonia synthesis and the Ostwald process for oxidizing ammonia into nitric acid.
Explanation:
To produce nitric acid (HNO₃) from nitrogen (N₂), hydrogen (H₂), and oxygen (O₂), we start with the Haber process to produce ammonia (NH₃) followed by the Ostwald process to oxidize ammonia into nitric acid. The net chemical equation represents a combination of these two processes.
The Balanced Equations for each step:
Haber process: N₂(g) + 3H₂(g) → 2NH₃(g)Ostwald process: 4NH₃(g) + 5O₂(g) → 4NO₂(g) + 6H₂O(l) followed by NO₂ + H₂O → HNO₃ + NO(g), and then finally, 3NO(g) + H₂O → 2HNO₃(l) + NO(g)Combining and cancelling out the common elements, we get the net balanced equation for the entire process:
N₂(g) + O₂(g) + 3H₂(g) → 2HNO₃(g) + 2H₂O(l)
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of wate
Answer: The final temperature of the mixture is 49°C
Explanation:
To calculate the mass of water, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
For cold water:Density of cold water = 1 g/mL
Volume of cold water = 230.0 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g[/tex]
For hot water:Density of hot water = 1 g/mL
Volume of hot water = 120.0 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g[/tex]
When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of hot water = 120 g
[tex]m_2[/tex] = mass of cold water = 230 g
[tex]T_{final}[/tex] = final temperature = ?°C
[tex]T_1[/tex] = initial temperature of hot water = 95°C
[tex]T_2[/tex] = initial temperature of cold water = 25°C
c = specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
[tex]120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)][/tex]
[tex]T_{final}=49^oC[/tex]
Hence, the final temperature of the mixture is 49°C
Can a metal and a nonmetal participate in a combination reaction
Answer:
Yes
Explanation:
A reaction normally takes place between metals and non metals. The metals acts as the electron donors and the non metals acts as the electron acceptors. This exchange of electrons form bonds such as ionic or covalent.
A good example of a reaction between a metal and non metal is Sodium metal and Chlorine(non metal). They form an ionic bond and the product is Sodium chloride.
A student may say hypothesis that the water on the outside of the paint evaporate faster than the inside . Describe an investigation of the student could use to test this hypothesis
Answer:
First of all, exterior paint is more resistant to extreme temperatures, humidity, pressure differences, the presence of droughts and this is due to its chemical composition, which is more resistant and because it penetrates the surface to be painted better (than in This case is the external wall of a home), by better penetrating the surface to which it is applied, there is less risk of detachment, of accumulating bubbles or air at the wall-paint interface, it adheres much better ... as explained this? Well, this phenomenon is due to the fact that it presents a solvent that helps to impregnate the wall (primer) and transports the paint as a vehicle to the internal pores of the wall.
On the other hand, in interior environments, the paints are less resistant to the aforementioned adverse factors and less impregnate the surfaces because the solvent is in less quantity or does not have the same impregnation capacity or essential primer as in the case of exterior paints. .
Explanation:
There are some specialists in the field who nowadays advise even interior paints with insulating or waterproof lacquers since the interior ones bear so little factors such as humidity that they no longer meet the basic requirements, that is why some interior paints they show fungus or wet stains against walls with wet content.
Answer:
Water on the outside of the paint evaporate faster than the inside. To test this hypothesis this student might place water in containers with different sizes and shapes in locations with different temperatures. This way it can be measured how fast water in different places evaporate.
Explanation:
A change in temperature can affect the physical properties of water. Water circulates through a continuous cycle. The sun’s energy drives weather patterns, and enables water to change through different stages in the water cycle : evaporation, condensations, precipitation, ground water, transpiration.
By placing water in containers with different sizes and shapes in locations with different temperatures, the student will see that the sun’s energy increases the amount of water that evaporates. More water evaporates when the temperature is warmer and when there is a large surface area or mouth/opening.
Match each exploration mission.
1. explored the Challenger Deep
Vostok
2. took pictures and video of Mars surface
Sputnik 1
3. first man-made object sent into space
Sojourner
4. first manned orbit around the earth
Apollo 11
5. first men to walk on the moon
Mariner 5
6. explored the dark side of Venus
Trieste i need help ASAP
Answer:
Match each exploration mission.
1. explored the Challenger Deep
5
Apollo 11
2. took pictures and video of Mars surface
3
Sputnik 1
3. first man-made object sent into space
2
Sojourner
4. first manned orbit around the earth
6
Trieste
5. first men to walk on the moon
4
Vostok
6. explored the dark side of Venus
1
Mariner 5
Explanation:
*
Answer:
Apollo 11 - First man to walk on the moon
Trieste - Explored the challenger deep
Sojourner - First man-made object sent into space
Vostok - First manned orbit around the earth
Mariner 5 - Explored the dark side of venus
Sputnik 1 - Took pictures and videos of Mars surface
Explanation:
A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and the anode solution has an unknown pH, has a cell voltage of 0.182 V. What is the pH in the unknown solution? Assume the temperature of the solutions is 298 K. pH =
The pH of the unknown solution is 3.07.
Explanation:
1.Find the cell potential as a function of pH
From the Nernst Equation:
Ecell=E∘cell−RT /zF × lnQ
where
R denotes the Universal Gas Constant
T denotes the temperature
z denotes the moles of electrons transferred per mole of hydrogen
F denotes the Faraday constant
Q denotes the reaction quotient
Substitute the values,
E∘cell=0 lnQ=2.303logQ
E0cell=−2.30/RT /zF × log Q
Solving the equation,
2. Find the Q value
Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant
Q=[H+]^2×1/1×1=[H+]2
Taking the log
logQ= log[H+]^2=2log[H+]=-2pH
From the formula,
Ecell=−2.303RT /zF× logQ
E cell= 2.303 × 8.314 CK mol (inverse) × 298.15
K × 2pH /2×96 485 C⋅mol
( inverse)
E cell= 0.0592 V × pH
3. Finding the pH value
E cell= 0.0592 V × pH
pH = E cell/ 0.0592 V= 0.182V/ 0.0592V
pH=3.07
The pH of the unknown solution is 3.07.
Calculate the pH for each case in the titration of 50.0 mL of 0.110 M HClO ( aq ) with 0.110 M KOH ( aq ) . Use the ionization constant for HClO . What is the pH before addition of any KOH ? pH = What is the pH after addition of 25.0 mL KOH ? pH = What is the pH after addition of 35.0 mL KOH ? pH = What is the pH after addition of 50.0 mL KOH ? pH = What is the pH after addition of 60.0 mL KOH ? pH =
pH before KOH: 4.23. After 25.0 mL KOH: 4.23. After 35.0 mL: 9.98. After 50.0 mL: 11.26. After 60.0 mL: 11.85.
To solve this problem, we'll use the stoichiometry of the reaction between HClO and KOH to find the amount of each reactant remaining at each point of the titration. Then, we'll use the relevant equilibrium equations to calculate the pH.
Step 1: Calculate the initial pH before adding any KOH.
[tex]\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \][/tex]
[tex]\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{K_a [\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{(3.5 \times 10^{-8})(0.110)} \][/tex]
[tex]\[ [\text{H}^+] \approx 5.92 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ \text{pH} = -\log{[\text{H}^+]} = -\log{(5.92 \times 10^{-5})} \][/tex]
[tex]\[ \text{pH} \approx 4.23 \][/tex]
Step 2: Calculate the pH after adding 25.0 mL of KOH.
Moles of HClO initially = [tex]\( 0.110 \, \text{M} \times 0.0500 \, \text{L} = 0.00550 \, \text{mol} \)[/tex]
Moles of KOH added = [tex]\( 0.110 \, \text{M} \times 0.0250 \, \text{L} = 0.00275 \, \text{mol} \)[/tex]
Remaining moles of HClO = [tex]\( 0.00550 \, \text{mol} - 0.00275 \, \text{mol} = 0.00275 \, \text{mol} \)[/tex]
Calculate the new concentration of HClO:
[tex]\[ \text{New volume of HClO} = 50.0 \, \text{mL} - 25.0 \, \text{mL} = 25.0 \, \text{mL} = 0.0250 \, \text{L} \][/tex]
[tex]\[ \text{New} \, [\text{HClO}] = \frac{0.00275 \, \text{mol}}{0.0250 \, \text{L}} \][/tex]
[tex]\[ \text{New} \, [\text{HClO}] = 0.110 \, \text{M} \][/tex]
Now, we use the equilibrium equation for the dissociation of HClO to find the concentration of [tex]\( \text{H}^+ \):[/tex]
[tex]\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \][/tex]
[tex]\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{K_a [\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{(3.5 \times 10^{-8})(0.110)} \][/tex]
[tex]\[ [\text{H}^+] \approx 5.92 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ \text{pH} = -\log{[\text{H}^+]} = -\log{(5.92 \times 10^{-5})} \][/tex]
[tex]\[ \text{pH} \approx 4.23 \][/tex]
The pH remains the same after adding 25.0 mL of KOH.
Let's now calculate the pH after adding 35.0 mL, 50.0 mL, and 60.0 mL of KOH.
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq) , as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 295 mL Cl2(g)295 mL Cl2(g) at 25 °C and 795 Torr795 Torr ?
Answer:
1.131g of MnO2
Explanation:
Step 1:
The balanced equation for the reaction.
MnO2(s) + 4HCl(aq)—> MnCl2(aq) + 2H2O(l) + Cl2(g)
Step 2:
Data obtained from the question. This includes:
Volume (V) of Cl2 = 295mL = 0.295L
Temperature (T) = 25°C = 25°C + 273 = 298K
Pressure (P) = 795 Torr = 795/760 = 1.05 atm
Number of mole (n) of Cl2 =.?
Gas constant (R) = 0.082atm.L/Kmol
Step 3:
Determination of the number of mole of Cl2 produced. The number of mole of Cl2 produced from the reaction can be obtained by applying the ideal gas equation as follow:
PV = nRT
1.05 x 0.295 = n x 0.082 x 298
Divide both side by 0.082 x 298
n = (1.05 x 0.295)/(0.082 x 298)
n = 0.013 mole.
Therefore, the number of mole of Cl2 produced is 0.013 mole.
Step 4:
Determination of number of mole of MnO2 needed to produce 0.013 mole of Cl2.
This is illustrated below:
From the balanced equation above,
1 mole of MnO2 produced 1 mole of Cl2.
Therefore, it will also take 0.013 mole of MnO2 to produce 0.013 mole of Cl2.
Step 5:
Converting 0.013 mole of MnO2 to grams. This is illustrated below:
Number of mole MnO2 = 0.013 mole
Molar Mass of MnO2 = 55 + (2x16) = 87g/mol
Mass of MnO2 =?
Mass = number of mole x molar Mass
Mass of MnO2 = 0.013 x 87
Mass of MnO2 = 1.131g
Therefore 1.131g of MnO2 should be added to excess HCl.
What evidence indicates that a reaction has occurred? (Select all that apply.)
a. The temperature decreased.
b. A solid brown product formed.
c. The temperature increased.
d. A gas formed.
e. An explosion occurred.
Final answer:
Evidence of a chemical reaction can be determined by changes such as temperature fluctuations, formation of a precipitate, gas production, and color changes. Temperatures changing, solid products forming, gas releasing, and explosions are all indications of chemical reactions.
Explanation:
Evidence indicating that a chemical reaction has occurred includes:
Temperature change: This can be an increase or a decrease, signifying an energy transfer during the reaction. Formation of a precipitate: A solid product forming in a previously clear solution is a tell-tale sign of a chemical change.Gas formation: The appearance of bubbles that are not the result of the substance boiling indicates a new gas is being produced.Color change: An unexpected change in color suggests that new substances with different properties are being formed.Based on these observations, the correct answers from the choices provided are: (a) The temperature decreased, (b) A solid brown product formed, (c) The temperature increased, (d) A gas formed, and potentially (e) An explosion occurred, as an explosion indicates a rapid energy transfer and creation of new substances.
There is a lot of talk about pH and shampoos. Some shampoos claim they balance the pH in your hair, others say they are alkaline and others say they are neutral or match the "natural" pH of your hair. What do you think that this means? Do you think the pH of your shampoo affects the condition of your hair? Why or why not? *
Answer:Yes it affects the conditions of the hair.
Explanation:most shampoos have pH values higher than the pH of the hair shaft of 3.6 and also higher than the pH of the hair scalp of 5.5. Therefore the usage of hair shampoo with pH values above 5.5 may likely increase the rate of friction which may cause hair breakage and facilitate hair tangling.
Shampoos with various pH ratings have different effects on hair. The pH level of your shampoo influences the health of your hair. Shampoos that match the hair's natural pH of 5 to 7 are typically best for maintaining hair health.
Explanation:The pH measurement, which ranges from 0 to 14, indicates whether a substance is acidic, neutral, or alkaline. Most shampoos usually have a pH level between 5 and 7, similar to hair's natural pH. When a shampoo claims to 'balance' the pH of your hair, it means it tries to keep this pH level, which is slightly acidic for a healthy scalp and hair. Meanwhile, neutral shampoos have a pH of 7, and alkaline shampoos have a pH higher than 7, which could potentially cause hair to become dry and brittle over time, especially with overuse. So, the pH of your shampoo does indeed affect your hair condition. Using a shampoo that aligns with the natural pH of your hair is generally considered optimal for maintaining hair health.
Learn more about shampoo pH here:https://brainly.com/question/32512053
#SPJ6
A chemist prepares a solution of potassium permanganate by measuring out 26. g of potassium permanganate into a 350. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium permanganate solution. Be sure your answer has the correct number of significant digits.
Answer:
0.471 mol/L
Explanation:
First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.
This is illustrated below:
Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol
Mass of KMnO4 from the question = 26g
Mole of KMnO4 =?
Number of mole = Mass/Molar Mass
Mole of KMnO4 = 26/158 = 0.165mole
Now we can obtain the concentration of KMnO4 in mol/L as follow:
Volume of the solution = 350mL = 350/1000 = 0.35L
Mole of KMnO4 = 0.165mole
Conc. In mol/L = mole of solute(KMnO4)/volume of solution
Conc. In mol/L = 0.165mol/0.35
conc. in mol/L = 0.471mol/L
Answer:
The concentration of this potassium permanganate solution is 0.470 mol/L or 0.470 M
Explanation:
Step 1: data given
Mass of potassium permanganate = 26.0 grams
Molar mass of potassium permanganate = 158.034 g/mol
Volume = 350 mL = 0.350 L
Step 2: Calculate moles potassium permanganate
Moles KMnO4 = mass KMnO4 / molar mass KMnO4
Moles KMnO4 = 26.0 grams / 158.034 g/mol
Moles KMnO4 = 0.1645 moles
Step 3: Calculate concentration of KMnO4 solution
Concentration KMnO4 = moles KMnO4 / volume
Concentration KMnO4 = 0.1645 moles / 0.350 L
Concentration KMnO4 = 0.470 M
The concentration of this potassium permanganate solution is 0.470 mol/L or 0.470 M