Answers
Answer: The monomer of nylon-6 is caprolactam.
Explanation:
From the question, we are asked to draw the cyclic monomer from the structure of Nylon-6.
Two repeating monomer units can be seen in the polymeric structure given in the problem.
From this we can see and conclude that The monomer of nylon-6 is caprolactam.
attached is a diagrammatic representation of the Structure (Nylon-6 monomer).
cheers i hope this helps!!!!
Final answer:
Nylon-6 is synthesized from the cyclic monomer epsilon-caprolactam through a process called ring-opening polymerization, unlike other nylons that are produced by condensation polymerization. Epsilon-caprolactam features a 6-carbon ring with an amide linkage, making it unique.
Explanation:
Nylon-6 is a unique type of polymer that is derived from a cyclic monomer known as epsilon-caprolactam. Unlike the Nylon 6,6 polymer, which is made from condensation polymerization of hexanedioic acid and 1,6-diaminohexane, Nylon-6 utilizes a process called ring-opening polymerization. This process involves breaking open the ring structure of the epsilon-caprolactam to form linear chains, which then link together head-to-tail to form the polymer. This is a key difference as it does not release a small molecule (like water or hydrogen chloride) during the polymerization process, making it a growth polymerization.
Epsilon-caprolactam, the monomer for Nylon-6, is a 6-carbon cyclic amide. To deduce and draw its cyclic monomer, one should recognize that the repeating unit in Nylon-6 consists of 6 carbons in a chain with an amide linkage connecting the ends. Thus, depicting epsilon-caprolactam involves drawing a ring structure containing 5 methylene (-CH2-) groups and one carbonyl group (=O) adjacent to an NH group, forming a lactam.
Related Questions
What solution is placed when plating metal with gold?
Answers
Answer:
The metal atoms that plate your object come from out of the electrolyte, so if you want to copper plate something you need an electrolyte made from a solution of a copper salt, while for gold plating you need a gold-based electrolyte—and so on
Explanation:
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Answer:
electrolyte which is made from a solution of copper salt
Explanation:
2K + 2HBr → 2 KBr + H2
When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)
●a). What is the limiting reactant?
●b.)What is the excess reactant?
●C.)How much product is produced?
Answers
Answer:
[tex]\large \boxed{\text{0.0503 g}}[/tex]
Explanation:
The limiting reactant is the reactant that gives the smaller amount of product.
Assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 39.10 80.41 2.016
2K + 2HBr ⟶ 2KBr + H₂
m/g: 5.5 4.04
a) Limiting reactant
(i) Calculate the moles of each reactant
[tex]\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}[/tex]
(ii) Calculate the moles of H₂ we can obtain from each reactant.
From K:
The molar ratio of H₂:K is 1:2.
[tex]\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}[/tex]
From HBr:
The molar ratio of H₂:HBr is 3:2.
[tex]\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}[/tex]
(iii) Identify the limiting reactant
HBr is the limiting reactant because it gives the smaller amount of NH₃.
b) Excess reactant
The excess reactant is K.
c) Mass of H₂
[tex]\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }[/tex]
b. If 20.0 grams of Aluminum and 30.0 grams of chlorine gasſare used, and how many
grams AlCl3 can theoretically be made (3 pts)?
Answers
Answer: The mass of [tex]AlCl_3[/tex] theoretically be made can be, 37.4 grams.
Explanation : Given,
Mass of [tex]Al[/tex] = 20.0 g
Mass of [tex]Cl_2[/tex] = 30.0 g
Molar mass of [tex]Al[/tex] = 27 g/mol
Molar mass of [tex]Cl_2[/tex] = 71 g/mol
First we have to calculate the moles of [tex]Al[/tex] and [tex]Cl_2[/tex].
[tex]\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{20.0g}{27g/mol}=0.741mol[/tex]
and,
[tex]\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}=\frac{30.0g}{71g/mol}=0.422mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
From the balanced reaction we conclude that
As, 3 moles of [tex]Cl_2[/tex] react with 2 moles of [tex]Al[/tex]
So, 0.422 moles of [tex]Cl_2[/tex] react with [tex]\frac{2}{3}\times 0.422=0.281[/tex] moles of [tex]Al[/tex]
From this we conclude that, [tex]Al[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Cl_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AlCl_3[/tex]
From the reaction, we conclude that
As, 3 moles of [tex]Cl_2[/tex] react to give 2 moles of [tex]AlCl_3[/tex]
So, 0.422 moles of [tex]Cl_2[/tex] react to give [tex]\frac{2}{3}\times 0.422=0.281[/tex] mole of [tex]AlCl_3[/tex]
Now we have to calculate the mass of [tex]AlCl_3[/tex]
[tex]\text{ Mass of }AlCl_3=\text{ Moles of }AlCl_3\times \text{ Molar mass of }AlCl_3[/tex]
Molar mass of [tex]AlCl_3[/tex] = 133 g/mole
[tex]\text{ Mass of }AlCl_3=(0.281moles)\times (133g/mole)=37.4g[/tex]
Therefore, the mass of [tex]AlCl_3[/tex] theoretically be made can be, 37.4 grams.
different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity from the results of laboratory analysis that the student could use to determine whether the sample was pure.
Answers
The student can check the purity of the CaBr2 sample by performing a quantitative chemical analysis to compare the theoretical and actual mass of the compound, which will reveal the presence of any impurities.
Explanation:A student can determine whether the 10.0g sample labeled CaBr2 is pure by calculating the molar mass of CaBr2 and comparing the theoretical mass to the actual mass of the sample. If there is a discrepancy between the theoretical and actual mass, it could indicate the presence of an inert impurity.
To carry out the analysis, the student must perform a quantitative chemical analysis, such as a titration, to determine the amount of active ingredient in the sample. The student could also measure the mass of CaBr2 after a precipitation reaction, as described in the problem statement. If the resulting mass corresponds to the known molar mass of a pure CaBr2 sample, then the sample can be considered pure. On the other hand, if the mass deviates from the expected value, then an impurity might be present.
An insoluble solid is placed in water and the system allowed to reach equilibrium. The ratio of the rate at which ions join the solution and the rate at which ions join the lattice will be:
a) greater than one
b) one
c) less than one
d) depends on the substance
Answers
Answer:
One
Explanation:
Because it reaches an equibrum state so it's equals to one
Why did hurricane Katrina slow down at data point 7 ?
Answers
Answer:
Because the hot air from the equator is balance with the cold air from the polar region, meaning the temperature is the right degree, therefore it causes the slowing down of that hurricane.
Explanation:
From your science class you do study the convectional current right? that's what happen on the outside real life
As more than just a result of the equator's hot air being in balance with the poles' cold air, which means the temperature is at the right level, the hurricane is slowed down.
What is hurricane ?Hurricanes are low-pressure systems with organized thunderstorm activity that originate over tropical or subtropical waters. They are sometimes referred to as tropical cyclones in general. The warm ocean waves provide them energy.
When warm, humid air begins to rise over water, hurricanes emerge. Cooler air replaces the rising air. Large clouds and thunderstorms continue to grow as a result of this process. Thanks to the earth's Coriolis Effect, these thunderstorms are still expanding and starting to rotate.
The greatest wind speeds during Katrina's impact near Grand Isle, Louisiana, may have reached 140 mph. The NWS Doppler Radar at Mobile (KMOB) recorded winds of up to 132 mph between 3,000 and 4,000 feet above ground level in the morning as Katrina moved farther north and made a second impact along the Mississippi/Louisiana border.
Thus, the equator's hot air being in balance with the poles' cold air, which means the temperature is at the right level, the hurricane is slowed down.
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Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reactions:
HSO3−(aq)+H2O(l)→H2SO3(aq)+OH−(aq)
(CH3)3N(g)+BCl3(g)→(CH3)3NBCl3(s)
Drag the appropriate items to their respective bins.
Answers
Answer:
1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)
The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻
2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)
There are no Brønsted-Lowry acids and bases in this reaction.
Explanation:
According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).
The chemical equation for this reaction is:
HA + B ⇌ A⁻ + HB⁺
Given reactions:
1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)
The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻
Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.
2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)
There are no Brønsted-Lowry acids and bases in this reaction.
Reason: In this reaction, there is no exchange of proton between the acid and the base.
Which of the following statements will decrease the amount of work the system could perform? (a) Add to the solution (b) Add solid NaOH to the reaction (assume no volume change) (c) Increase the concentration of the (d) Selectively remove (e) Add to the solution
Answers
Final answer:
The addition of solid NaOH to a reaction can decrease the amount of work a chemical system can perform by shifting the equilibrium position and altering the concentration of reactants and products involved in work
Explanation:
The statement that will decrease the amount of work the system could perform is (b) Add solid NaOH to the reaction (assume no volume change). In a chemical system, work is related to changing conditions that affect the direction of equilibrium according to Le Chatelier's principle. Adding solid NaOH to a reaction mixture could shift the equilibrium position in a way that may result in the consumption of reactants or products involved in performing work (such as electrical or mechanical work in an electrochemical cell or in muscle contraction). For example, if the reaction were exergonic and NaOH was a product, adding more of it would shift the equilibrium to the left, thereby reducing the amount of work the system could do. This is because there would be a greater proportion of reactants and less of the energy-containing products required to perform work.
Which of the following is one way that nuclear reactions differ from chemical
reactions?
Answers
Answer:
Nuclear reaction takes place at the nucleus whereas chemical reaction involves valence electrons
Explanation:
If the hydroxide ion concentration is 10-7 M, what is the pH of the solution
Answers
Answer:
pH= 7
Explanation:
pOH=-log [OH-] = -log (10^-7)= 7
pH= 14-pOH= 14-7
=7
The shells of marine organisms contain calcium carbonate CaCO3, largely in a crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical and thermodynamic properties of calcite and aragonite are given below.
Properties
(T=298K, P=1atm) Calcite Aragonite
ΔH°f (kJ/mole) -1206.87 -1207.04
ΔG°f (kJ/mole) -1128.76 -1127.71
S° (J/mole K) 92.88 88.70
C°P (J/mole K) 81.88 81.25
Density (gm/mL) 2.710 2.930
A) Based on the thermodynamic data given, would you expect an isolated sample of
calcite at T=298K and P=1 atm to convert to aragonite, given sufficient time.
Explain.
B) What pressure must be achieved to induce the conversion of calcite to aragonite at
T=298K. Assume both calcite and aragonite are incompressible at T=298K.
C) Suppose the pressure is P=1.00 atm and the temperature is increased from
T=298K to T=400K. Based on this calculation, can isolated calcite be converted
to aragonite at P=1 atm if the temperature is increased? Explain.
D) Suppose the pressure is increased from P=1.00 atm to P=1000 atm and the
temperature is increased from T=298K to T=400K. Which form of CaCO3 is more
stable under these conditions?
Expert Answer
Answers
Answer:
Check the explanation
Explanation:
A. Calcite ------> Argonite
At 298 K and 1 atm,
\DeltaGrxn = \DeltaGarg - \DeltaGcalc
= -1127.71-(-1128.76) = 1.05 kJ/mol
\DeltaGrxn is positive, so the reaction is not spontaneous.
Thus, the reaction would not happen even if enough time is given.
B. Molar mass of CaCO3 = 100 g/mol
Volume = mass/density
Vcalc = 100/2.710 = 36.900 ml/mol = 36.9*10-3 L = 36.9*10-6 m3
Varg = 100/2.930 = 34.130 ml/mol = 34.13*10-3 L = 34.13*10-6 m3
\DeltaGrxn = 1.05 kJ/mol = 1.05*103 J/mol
\DeltaG = -\DeltaP\DeltaV
\DeltaP = -\DeltaG/\DeltaV = -1.05*103/(34.13*10-6 - 36.9*10-6) = 0.379*109 Pa = 3.79*108 Pa
The pressure should be more than 3.79*108 Pa.
C. For reaction, \DeltaSrxn = Sarg - Scalc
= 88.70-92.88 = -4.18 J/mol.K
\DeltaHrxn = -1207.04-(-1206.87) = -0.17kJ/mol
At 400 K, \DeltaG = \DeltaH-T\DeltaS
= -0.17-(400*-4.18)
= +ve
So, the reaction is not favorable on increasing the temperature to 400 K.
D. \DeltaT = 400-298 = 102 K
\DeltaP = 1000-1 = 999.0 atm = 999*106 Pa
\DeltaG = \DeltaS\DeltaT - \DeltaV\DeltaP
= (-4.18*102)-(-2.77*10-6*999*106)
= -426.36+2767.23 = 2340.87 J/mol
\DeltaG = +ve
calcite would be more stable.
When The reaction below demonstrates which characteristic of a base?
Upper C U upper S upper O subscript 4 (a q) plus 2 upper N a upper O upper H (a q) right arrow upper C u (upper O upper H) subscript 2 (s) plus upper N a subscript 2 upper S upper O subscript 4 (a q).does phenolphthalein turn pink?
Answers
It shows the neutralization property of the base and there is no pink color.
Explanation:
CuSO₄(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
Here NaOH is the base and Copper sulfate is an salt of an acid and so it is acidic salt. When these two compounds react, we will get the precipitate of copper hydroxide and Sodium sulfate.
And this reaction seems to be a neutralization reaction, since an acidic salt is neutralized by a base.
Since NaOH is converted into salt, that is sodium sulfate (Na₂SO₄), the phenolphthalein indicator becomes colorless in this reaction. It doesn't turn pink.
Determine the amount of heat energy in joules required to raise the temperature of 7.40g of water from 29.0OC to 46.0 OC.
Answers
Answer: The amount of heat energy in joules required to raise the temperature is 526 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed = ?
m= mass of substance = 7.40 g
c = specific heat capacity = [tex]4.184J/g^0C[/tex]
Initial temperature of the water = [tex]T_i[/tex] = 29.0°C
Final temperature of the water = [tex]T_f[/tex] = 46.0°C
Change in temperature ,[tex]\Delta T=T_f-T_i=(46.0-29.0)^0C=17.0^0C[/tex]
Putting in the values, we get:
[tex]Q=7.40g\times 4.184J/g^0C\times 17.0^0C[/tex]
[tex]Q=526J[/tex]
The amount of heat energy in joules required to raise the temperature is 526 Joules
The amount of heat energy will be "526 J".
Given:
Mass,
m = 7.40 gInitial temperature,
[tex]T_i[/tex] = 29.0°CFinal temperature,
[tex]T_f[/tex] = 46.0°CChange in temperature,
[tex]\Delta T = T_f -T_i[/tex][tex]= 46.0-29.0[/tex]
[tex]= 17.0^{\circ} C[/tex]
The amount of heat energy,
→ [tex]Q = mc \Delta T[/tex]
By substituting the values, we get
[tex]= 7.40\times 4.184\times 17.0[/tex]
[tex]= 526 \ J[/tex]
Thus the above answer is right.
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Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochemical series for ligands and metals? (Select one answer. There is only one correct answer)
a. Strong o-donor ligands raise the energy of the eg orbitals of an octahedral complex.
b. Higher oxidation state metals form stronger bonds with ligands.
c. Ligands such as CO, PPhy, and bipyridine can act as T-acids.
d. The 4d and 5d transition metals are stronger Lewis acids than the 3d transition metals.
e. The electronegativity of the donor atom is the most important factor for ligands
Answers
Answer:
Higher oxidation state metals form stronger bong with ligands
Explanation:
Ligand strength are based on oxidation number, group and its properties
Statement e is not an accurate description because the ordering of the spectrochemical series depends on multiple factors, including ligand size, pi-donor or pi-acceptor abilities, and the metal's identity and oxidation state, in addition to electronegativity of the donor atom.
The correct answer to the question "Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochemical series for ligands and metals?" is e. The electronegativity of the donor atom is the most important factor for ligands.
The spectrochemical series is an ordering of ligands based on the magnitude of the ligand field splitting parameter (Δoct) they produce in coordination complexes. Contrary to option e, it is not just the electronegativity of the donor atom but also a series of other factors such as the size of the donor atom, the ligand's field strength (pi-donor or pi-acceptor abilities), and the identity of the metal ion and its oxidation state that contribute to the ordering of the spectrochemical series. For example, small neutral ligands with highly localized lone pairs, such as NH₃, produce larger Δoct values which illustrates more than just electronegativity playing a role.
Maria needs to dilute the stock of 16.0 M HCl solution for a lab.
How many mL of water must she add to 5.00 mL of 16.0 M HCI
solution to prepare a 1.00 M HCl solution? __ mL
Answers
Answer: 80mL
Explanation:
Its on ck12
5*16=80
To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to add 75 mL of water to the 5.00 mL of the stock solution.
Explanation:To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to dilute the stock. She should add water to the 5.00 mL of the stock solution. Let's calculate the volume of water she needs to add.
First, we can use the equation (stock concentration) × (stock volume) = (final concentration) × (final volume) to find the volume of the diluted solution. Plugging in the values, we have (16.0 M) × (5.00 mL) = (1.00 M) × (final volume). Solving for the final volume, we get:
final volume = (16.0 M × 5.00 mL) / 1.00 M
final volume = 80 mL
To find the volume of water Maria needs to add, we subtract the volume of the stock solution from the final volume:
volume of water = final volume - volume of stock solution = 80 mL - 5.00 mL = 75 mL
Therefore, Maria needs to add 75 mL of water to the 5.00 mL of 16.0 M HCl solution to prepare a 1.00 M HCl solution.
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Calculate the amount of heat needed to melt 148. g of solid octane (C8H18 ) and bring it to a temperature of 117.8 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answers
Answer:
The amount of heat is 84.4894kJ
Explanation:
Given data:
mass of octane=148g
The moles of octane is:
[tex]n_{octane} =148g*\frac{1mol}{114.23g} =1.2956moles[/tex]
The melting point is=-56.82°C
ΔHfus=20.73kJ/mol
The heat of fusion is:
[tex]Q_{fus} =n_{octane} *delta-H_{fus} =1.2956*20.73=26.8578kJ[/tex]
The heat to bring the octane to a temperature of 117.8°C is:
[tex]Q_{2} =mCp(117.8-(56.82))=148*2.23*(117.8+56.82)=57631.5848J=57.6316kJ[/tex]
The total heat in the process is:
Qtotal=Qfus+Q₂=26.8578+57.6316=84.4894kJ
A proposed mechanism is: Step 1: H2(g) + ICl(g) → HI(g) + HCl(g) (slow) Step 2: HI(g) + ICl(g) → I2(g) + HCl(g) (fast) Which of the following species is a catalyst? ICl HCl This mechanism has no catalyst. HI H2
Answers
Answer: This mechanism has no catalyst
Explanation:
The proposed mechanism is :
Step 1 : [tex]H_2(g)+ICl(g)\rightarrow HI(g)+HCl(g)[/tex] (slow)
Step 2: [tex]HI(g)+ICl(g)\rightarrow I_2(g)+HCl(g)[/tex] (fast)
The combined chemical equation will be :
[tex]H_2(g)+2ICl(g)\rightarrow I_2(g)+2HCl(g)[/tex]
A catalyst is a substance which enhances the rate of chemical reaction without being consumed in the chemical reaction. Thus catalyst gets used up in first step and gets regenerated in second.
HI is formed as an intermediate as it gets formed in first step and gets used up in the second step.
Here there is no substance which is used as a catalyst.
Can 1750 mL of water dissolve 4.6 moles of copper sulfate CuSO4
Answers
Answer:
No
Explanation:
We first find the solubility of the copper sulphate salt
Number of moles= 4.6 moles
Volume =1.75 L
Molarity= number of moles/ volume = 4.6/1.75= 2.6 molL-1
Hence, only 2.6 moles of Copper II sulphate dissolves in 1.75L of water. Hence 4.6 moles of copper II Sulphate dies not dissolve in 1.75 L of water.
list the 5 main factors that can affect the rates of a chemical reaction
Answers
Answer:
1.surface area of a solid reactant.2.concentration or pressure of a reactant.
3temperature.
4nature of the reactants.
5presence/absence of a catalyst.
Explanation:
The five main factors that can affect rates of a chemical reaction are concentration of reactants, temperature, pressure, presence of a catalyst, and surface area of solid reactants.
Explanation:The rate of a chemical reaction can be influenced by several factors, including:
Concentration of reactants: A higher concentration typically leads to an increased reaction rate because there are more molecules available to react.Temperature: Increasing the temperature often increases the reaction rate. This happens because higher temperatures give molecules more kinetic energy, so they collide more frequently and more forcefully.Pressure: Increasing the pressure of a gas reaction will increase the reaction rate because it effectively increases the concentration of the reactants.Presence of a catalyst: Catalysts lower the energy barrier for a reaction, making it easier for the reaction to proceed and thus increasing the reaction rate.Surface area of solid reactants: A larger surface area allows for more collisions between molecules, leading to faster reactions.Learn more about Chemical Reaction Rate Factors here:https://brainly.com/question/34204565
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How many grams of carbon dioxide are created from the complete combustion of 21.3 L of butane at STP?
Answers
Answer:
165.5 g of CO2
Explanation:
We must first put down the balanced reaction equation:
C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H2O(g)
From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:
22.4 L of butane created 174 g of CO2
Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2
Warning signs are sometimes placed on aerosol cans to prevent people from throwing them into a fire. What would be true about the contents of the gas in an aerosol can just after it is placed in a fire?
Answers
Answer:
Volume stays.
Explanation:
If aerosol can throwing into the fire than the temperature of the aerosol can & the temperature of the contents of the gas in an aerosol will also increase, and it induces the pressure to rise.
Against the sides of the can, the gas molecules will smash rapidly with each other which are present inside an aerosol can, and the proportion of gas will stay the same as earlier. According to the law of conservation of matter, as a result of the heat, the gas particles can not be eradicated or expanded.
From this combustion equation, 2CH22 + 3102 - 22H,0 + 2000, calculate the liters of
carbon dioxide produced when 16.9 grams of CH are combusted
Answers
Answer : The volume of [tex]CO_2[/tex] produced are, 26.7 liters.
Explanation :
First we have to calculate the moles of [tex]C_{10}H_{22}[/tex]
[tex]\text{Moles of }C_{10}H_{22}=\frac{\text{Given mass }C_{10}H_{22}}{\text{Molar mass }C_{10}H_{22}}=\frac{16.9g}{142g/mol}=0.119mol[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The given combustion reaction is:
[tex]2C_{10}H_{22}+31O_2\rightarrow 22H_2O+20CO_2[/tex]
From the balanced chemical reaction we conclude that,
As, 2 moles of [tex]C_{10}H_{22}[/tex] react to give 20 moles of [tex]CO_2[/tex]
So, 0.119 moles of [tex]C_{10}H_{22}[/tex] react to give [tex]\frac{20}{2}\times 0.119=1.19[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex] produced.
As we know that, 1 mole of gas occupies 22.4 L volume of gas.
As, 1 mole of [tex]CO_2[/tex] gas occupies 22.4 L volume of [tex]CO_2[/tex] gas.
So, 1.19 mole of [tex]CO_2[/tex] gas occupies 1.19 × 22.4 L = 26.7 L volume of [tex]CO_2[/tex] gas.
Therefore, the volume of [tex]CO_2[/tex] produced are, 26.7 liters.
17.131) A 24.0 −mL volume of a sodium hydroxide solution requires 19.5 mL of a 0.193 M hydrochloric acid for neutralization. A 11.0 −mL volume of a phosphoric acid solution requires 34.8 mL of the sodium hydroxide solution for complete neutralization. Calculate the concentration of the phosphoric acid solution
Answers
Answer:
The concentration of Phosphoric acid required for the neutralization described = 0.165 M
Explanation:
Given,
Volume of NaOH = V = 24.0 mL
Concentration of HCl = Cₐ = 0.193 M
Volume of HCl = Vₐ = 19.5 mL
NaOH + HCl -----> NaCl + H₂O
1 mole of NaOH reacts with 1 mole of HCl
Using the equivalence point expression
(CₐVₐ)/(CV) = (nₐ/n)
where
Cₐ = concentration of acid = 0.193 M
Vₐ = volume of acid = 19.5 mL
C = concentration of base = ?
V = volume of base = 24.0 mL
nₐ = Stoichiometric coefficient of acid in the balanced equation = 1
n = Stoichiometric coefficient of base in the balanced equation = 1
(CₐVₐ)/(CV) = (nₐ/n)
(0.193 × 19.5)/(C × 24) = 1
(C × 24) = 3.7635
C = (3.7635/24) = 0.157 M
This NaOH is then reacted with phosphoric acid.
Phosphoric acid = H₃PO₄
3NaOH + H₃PO₄ --------> Na₃PO₄ + 3H₂O
3 moles of NaOH reacts with 1 mole of Phosphoric acid.
Using the equivalence point expression
(CₐVₐ)/(CV) = (nₐ/n)
where
Cₐ = concentration of acid = ?
Vₐ = volume of acid = 34.8 mL
C = concentration of base = 0.157 M
V = volume of base = 11.0 mL
nₐ = Stoichiometric coefficient of acid in the balanced equation = 1
n = Stoichiometric coefficient of base in the balanced equation = 3
(CₐVₐ)/(CV) = (nₐ/n)
(Cₐ × 11)/(0.157 × 34.8) = (1/3)
Cₐ × 11 × 3 = 0.157 × 34.8 × 1
33Cₐ = 5.457
Cₐ = (5.457/32)
Cₐ = 0.165 M
Hence, the concentration of Phosphoric acid required for the neutralization described = 0.165 M
Hope this Helps!!!
How many molecules are in 23 Moles of oxygen o2
Answers
Answer:
1.38×10^25 molecules
Explanation:
Applying n= (no. of molcules)/NA
23 = N/6.02×10^23
= 1.38×10^25 molecules
What temperature must be maintained to ensure that a 1.00 L flask containing 0.0400 mol of oxygen will show a continuous pressure of 0.981 atm?
Answers
Answer:
The temperature is 298.9K = 25.75 °C
Explanation:
Step 1: Data given
Volume of the flask = 1.00 L
Number of moles oxygen = 0.0400 moles
Pressure = 0.981 atm
Step 2: Calculate the temperature
p*V = n*R*T
⇒with p = the pressure of oxygen gas = 0.981 atm
⇒with V = the volume of the flask = 1.00 L
⇒with n = the number of moles of oxygen = 0.0400 moles
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temeprature = TO BE DETERMINED
T = (p*V) / (n*R)
T = (0.981 atm * 1.00 L) / (0.0400 moles * 0.08206 L*atm/mol*K)
T = 298.9 K
The temperature is 298.9K = 25.75 °C
In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y 4 − Y4− ) and metal chelate (abbreviated MY n − 4 MYn−4 ) can buffer the free metal ion concentration at values near the dissociation constant of the metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant. The equilibrium M n + + Y 4 − − ⇀ ↽ − MY n − 4 Mn++Y4−↽−−⇀MYn−4 is governed by the equation K ′ f = α Y 4 − ⋅ K f = [ MY n − 4 ] [ M n + ] [ EDTA ] Kf′=αY4−⋅Kf=[MYn−4][Mn+][EDTA] where K f Kf is the association constant of the metal and Y 4 − Y4− , α Y 4 − αY4− is the fraction of EDTA in the form Y 4 − Y4− , and [ EDTA ] [EDTA] is the total concentration of free (unbound) EDTA EDTA . K ′ f Kf′ is the conditional formation constant. How many grams of Na 2 EDTA ⋅ 2 H 2 O Na2EDTA⋅2H2O (FM 372.23 g/mol) should be added to 1.97 1.97 g of Ba ( NO 3 ) 2 Ba(NO3)2 (FM 261.35 g/mol 261.35 g/mol ) in a 500. mL volumetric flask to give a buffer with p Ba 2 + = 7.00 pBa2+=7.00 at pH 10.00? log K f logKf for Ba − EDTA Ba−EDTA is 7.88 7.88 and α Y 4 − αY4− at pH 10.00 is 0.30. mass Na 2 EDTA ⋅ 2 H 2 O.
Answers
Final answer:
To prepare the buffer, you would need to add 2.808 grams of Na2EDTA · 2H2O to the 500 mL volumetric flask containing 1.97 grams of Ba(NO3)2. This will give the desired buffer with a pBa2+ of 7.00 at pH 10.00.
Explanation:
To calculate the mass of Na2EDTA · 2H2O needed to prepare the buffer, we need to use the equation:
Kf' = αY4− · Kf = [MYn−4][Mn+][EDTA]
Given that log Kf = 7.88 and αY4− = 0.30, we can rearrange the equation to solve for the concentration of MYn−4:
[MYn−4] = Kf' / (αY4− · Kf) = 10^(7.00 − 7.88) / (0.30 · 10^7.88) = 0.562 M
Now, we can calculate the moles of Ba(NO3)2:
moles of Ba(NO3)2 = mass / molar mass = 1.97 g / 261.35 g/mol = 0.00754 mol
Since the stoichiometric ratio between Na2EDTA · 2H2O and Ba(NO3)2 is 1:1, the moles of Na2EDTA · 2H2O required is also 0.00754 mol.
Finally, we can calculate the mass of Na2EDTA · 2H2O:
mass = moles × molar mass = 0.00754 mol × 372.23 g/mol = 2.808 g
Identify each of these substances as acidic, basic, or neutral.
rainwater, pOH = 8.5
cola, pOH = 11
tomato juice, POH = 10
liquid drain cleaner, poH = 0
Answers
Answer:
Rain Water: Acidic
Cola: Acidic
Tomato Juice: Acidic
Liquid Drain Cleaner: Basic
Rain Water: Acidic
Cola: Acidic
Tomato Juice: Acidic
Liquid Drain Cleaner: Basic
pHIf the value of pH is 7, then the solution is neutral, if greater than 7 then basic, and if less than 7 then acidic.
The relation between pH and pOH is as follows:-
[tex]pH+pOH=14[/tex]
The value of pH for rain water is 14-8.5=5.5.
So, the rain water is acidic.
The value of pH for cola is 14-11=3.
So, the cola is acidic.
The value of pH for tomato juice is 14-10=4.
So, the tomato juice is acidic.
The value of pH for liquid drain cleaner is 14-0=14
So, the liquid drain cleaner is basic.
Find more information about the pH here,
brainly.com/question/15289741
The common laboratory solvent diethyl ether (ether) is often used to purify substances dissolved in it. The vapor pressure of diethyl ether , CH3CH2OCH2CH3, is 463.57 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 7.745 grams of the compound were dissolved in 159.9 grams of diethyl ether, the vapor pressure of the solution was 457.87 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound?
Answers
Answer: The molecular weight of this compound is 288.4 g/mol
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : 7.745 g of compound is present in 159.9 g of diethyl ether
moles of solute = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.745g}{Mg/mol}[/tex]
moles of solvent (diethyl ether) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{159.9g}{74.12g/mol}=2.157moles[/tex]
Total moles = moles of solute + moles of solvent = [tex]\frac{7.745g}{Mg/mol}+2.157[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}[/tex]
[tex]\frac{463.57-457.87}{463.57}=1\times \frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}[/tex]
[tex]M=288.4g/mol[/tex]
Thus the molecular weight of this compound is 288.4 g/mol
The acids and bases shown right cover a range of pH values. Use what you know about acids, bases, and concentration to label the test tubes, in order, from most acidic to most basic.
Answers
Answer:
.1m HCl 0.001HCl .00001m HCl Distilled water .00001NaOH .001NaOH .01NaOH
Explanation:
Thanks to the person who posted the answers, I just wrote it out so its easier to see. :D
In the given compounds, the most acidic compound is 0.1m HCl and most basic compound is 0.01NaOH.
What are acids and bases?
Acids are those compounds whose has a pH value in the range from 0 to 7 and bases are those compounds which has a pH range from 7 to 14.
pH of any solution will be calculated as:
pH = -log[H⁺], where
[H⁺] = concentration of H⁺ ion and this concentration is present in the form of molarity (molar concentration).
So, pH value is directly proportional to the concentration value of H⁺ ion as concentration of H⁺ ion decreases so acidity also decreases and sequence of given compounds from most acidic to most basic is represented as:
0.1m HCl > 0.001HCl > 0.00001m HCl > Distilled water > 0.00001NaOH > 0.001NaOH > 0.01NaOH.
Hence 0.1m HCl is most acidic and 0.01 NaOH is most basic.
To know more about acidity & basicity, visit the below link:
https://brainly.com/question/940314
At 823 °C, Kp = 490 for the equilibrium reaction CoO(s) + CO(g) ⟷ Co(s) + CO2(g) What is the value of Kc at the same temperature for the equilibrium below? Co(s) + CO2(g) ⟷ CoO(s) + CO(g)
Answers
The equilibrium constant Kc for the reaction Co(s) + [tex]CO_{2}[/tex](g) ----> CoO(s) + CO(g) at 823°C is the reciprocal of Kp for the reverse reaction, resulting in a Kc of approximately 2.04 x 10^-3.
The value of the equilibrium constant Kc for the reaction Co(s) + [tex]CO_{2}[/tex](g) --------> CoO(s) + CO(g) can be calculated using the relationship between Kp and Kc and the fact that the reaction is the reverse of the one given. For the given reaction, Kp = 490 at 823°C. Since there are the same number of moles of gas on each side of the reaction, the relationship between Kp and Kc does not require any correction for the change in moles of gas, so Kc is the reciprocal of Kp for the reverse reaction. Thus, Kc for Co(s) + [tex]CO_{2}[/tex] ------->CoO(s) + CO(g) is 1/490 or approximately 2.04 x 10^-3 at 823°C.