The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.
The rate law is: rate = 24.00 × 10²⁵ M/s
The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t
If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.
The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.
Let's consider the decomposition of ethanol on an alumina surface.
C₂H₅OH(g) ⇒ C₂H₄(g) + H₂O(g)
The plot of [A] vs time (t) resulted in a straight line, which indicates that the reaction follows zero-order kinetics.
The slope, 24.00 × 10²⁵ M/s, represents the rate constant, k.
What is zero-order kinetics?It is a chemical reaction in which the rate of reaction is constant and independent of the concentration of the reacting substances
The rate law for zero-order kinetics is:
rate = 24.00 × 10²⁵ M/s
The integrated rate law for zero-order kinetics is:
[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t
What is the half-life?Is the time for the amount of substance to decrease by half.
If the initial concentration of C₂H₅OH was 1.25 × 10²² M, we can calculate the half-life [t(1/2)] using the following formula.
t(1/2) = [C₂H₅OH]₀ / 2 × k
t(1/2) = (1.25 × 10²² M) / 2 × (24.00 × 10²⁵ M/s) = 2.60 × 10⁻⁵ s
We can calculate the time required for all the 1.25 × 10²² M C₂H₅OH to decompose using the integrated rate law.
[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t
0 M = 1.25 × 10²² M - 24.00 × 10²⁵ M/s × t
t = 5.21 × 10⁻⁵ s
The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.
The rate law is: rate = 24.00 × 10²⁵ M/s
The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t
If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.
The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.
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The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul
Answer:
a) [tex]V_1=5ul[/tex]
b) [tex]v=20ul[/tex]
Explanation:
From the question we are told that:
initial Concentration [tex]C_1=50mg/ml[/tex]
Final Concentration [tex]C_2=10mg/ml[/tex]
Final volume needs [tex]V_2 =25ul[/tex]
Generally the equation for Volume is mathematically given by
[tex]C_1V_1=C_2V_2[/tex]
[tex]V_1=\frac{C_1V_1}{C_2}[/tex]
[tex]V_1=\frac{10*25}{50}[/tex]
[tex]V_1=5ul[/tex]
Therefore
The volume of buffer needed is
[tex]v=V_2-V_1\\\\v=25-5[/tex]
[tex]v=20ul[/tex]
Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. physical change d. element f. chemical change
1. homogenous: sugar solution
2. heterogeneous: sand solution
3. compound: water
4. physical change: ice melting
5. element: hydrogen
6. chemical change: burning fire
Based on the standard EMF series and your knowledge of half-reactions, determine the cell potential and spontanei ty of a cell that consists of a pure cobalt electrode in a solution of Co^2+ ions; the other half is a lead electrode immersed in a Pb^2+ solution.
Pb +2e- Pb Sn +2e Sn Ni 2e Ni Co 2e -0.126 -0.136 -0.250 -0.277 Co
a. +0.403, spontaneous
b. -0.403, nonspontaneous
c. +0.151, spontaneous
d. -0.151, nonspontaneous
Answer:
+0.151, spontaneous
Explanation:
Given that;
Co^2+(aq) + 2e ---->Co(s) -0.28 V
Pb^2+(aq) + 2e ---->Pb(s). -0.13 V
Hence Co is the anode and Pb is the cathode
E°cell = E°cathode - E°anode
So;
E°cell = -0.13 V - (-0.28 V)
E°cell = 0.15 V
The cell reaction is spontaneous since E°cell is positive.
what is the IUPAC name of 2NaOH(s)
Answer:
NaoH= sodium hydroxide
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
From the calculation, the mass of the product is 2 g.
What is a reaction?A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.
Number of moles of Na2CO3 = 100/1000 L * 0.2 mol/L = 0.02 moles
Number of moles of Ca(NO3)2 = 200/1000 L * 0.1 mol/L = 0.02 moles
Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g
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Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
How is the compound NH3 classified?
A. As a salt
B. As a base
C. As an acid
D. As ionic
Answer:
B
Explanation:
Ammonia is considered a base as it's pH is 11
Answer from Gauthmath
The compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.
What is Ammonia (NH3)?Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.
Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.
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Given its formula and Avogadro's Number (6.02 x 10^23 molecules/mol), deduce how many molecules are present in 3 x 10^-16 grams of TCDD. Type in only a number without using scientific notation.
Answer:
5 × 10⁵ molecules (500,000 molecules)
Explanation:
Step 1: Convert 3 × 10⁻¹⁶ g to moles
We will use the molar mass of TCDD (321.97 g/mol).
3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol
Step 2: Convert 9 × 10⁻¹⁹ mol to molecules
The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).
9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules
Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce 0.166 g of water and 0.403 g of carbon dioxide. What is the empirical formula for valproic acid
Answer:
The empirical formula is C4H8O
Explanation:
Step 1: Data given
Valproic acid is composed of C, H, and O
Mass of the sample = 0.165 grams
Mass of water = 0.166 grams
Mass of CO2 = 0.403 grams
Molar mass of water ( H2O) = 18.02 g/mol
Molar mass of CO2 = 44.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Atomic mass H = 1.01 g/mol (H2 = 2.02 g/mol)
Step 2: The equation
CxHyOz + O2 → CO2 + H2O
Step 3: Calculate the number of carbon in the sample
The carbon comes from CO2
Mass C = (12.01 g/mol/44.01 g/mol) * 0.403 grams
Mass C = 0.110 grams
Step 4: Calculate mass of hydrogen in the sample
The hydrogen comes from H2O
Mass H = (2.02/18.02) * 0.166 grams
MAss H = 0.0186 grams
Step 5: Calculate mass of O
The mass of O in the sample = Mass of sample - mass of H - mass of C
The mass of O = 0.165 grams - 0.110 grams - 0.0186 grams
The mass of O = 0.0364 grams
Step 6: Calculate moles
Moles C = 0.110 grams / 12.01 g/mol = 0.00916 moles
Moles H = 0.0186 / 1.01 = 0.0184 moles
Moles O = 0.0364/16.0 = 0.00228 moles
Step 7: Calculate empirical formula
We divide by the smallest amount of moles
C: 0.00916/ 0.00228 = 4
H: 0.0184/0.00228 = 8
O: 0.00228/0.00228 = 1
The empirical formula is C4H8O
State whether the error introduced by each of the following problems would result in a high or a low value for the Cu recovery or would not affect the results. Explain. a. Some of the copper nitrate solution splashes out of the beaker in step 1. _____________
Answer:
Low value for copper recovery
Explanation:
The percentage recovery is obtained from;
Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100
Note that the fact that some of the copper nitrate solution splashed out of the beaker means that some amount copper has been lost from the system. This loss of copper leads to a lower value of copper recovered from solution.
How many moles of Al are needed to react exactly with 10.00 moles of Fe2O3 according to the following
equation?
Fe2O3 + 2 Al → Al2O3 + 2Fe
A) 15.0 moles
1
B) 20.0 moles
C) 30.0 moles
D) 60.0 moles
E) 35.0 moles
Answer:
Answer is B) 20.0 moles
Explanation:
From the equation,
1 mole of Fe2O3 = 2 moles of Al
therefore 10.0 moles of Fe2O3 = 10×2
= 20.0 moles.
How many nitrate ions are present in the following aqueous solution?
5.08 L of a solution containing 2.68 x 1021 formula units of lithium nitrate per liter.
When you have the number, determine its log (base 10) and enter that value with 3 decimal places
Answer:
22.1
Explanation:
Step 1: Calculate the number of formula units of lithium nitrate
A 5.08 L solution contains 2.68 × 10²¹ formula units per liter.
5.08 L × 2.68 × 10²¹ formula units/1 L = 1.36 × 10²² formula units
Step 2: Calculate the number of nitrate ions
Lithium nitrate dissociates completely in water according to the following equation.
LiNO₃(aq) ⇒ Li⁺(aq) + NO₃⁻(aq)
The molar ratio of LiNO₃ to NO₃⁻ is 1:1. The number of nitrate ions is 1/1 × 1.36 × 10²² = 1.36 × 10²².
Then,
log (1.36 × 10²²) = 22.1
The number of nitrate ions are present in the following aqueous solution is 13.6x10²¹, and log(base 10) value of this is 22.2.
What are strong electrolytes?
Those salts which are completely dissociate into their ions in the solution form will known as strong electrolyte.
Lithium nitrate is a strong electrolyte and it shows complete dissociation as:
LiNO₃ → Li⁺ + NO₃⁻
Given that,
1 liter of solution containing = 2.68 x 10²¹ formula unit
5.08 L of solution containing = 5.08 x 2.68 x 10²¹ = 13.6x10²¹ formula unit
So, number of nitrate ion in given solution = 13.6x10²¹ formula unit
Log(13.6x10²¹) = 22.1
Hence, required values are 13.6x10²¹ and 22.1.
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When the equation,
O2 + __C 10H 22 →
CO2 +
H2O is balanced, the coefficient of O2 is?
Please help!
b. Sodium has an emission spectrum with two visible wavelengths, both very close to 590 nm. If you had a light source that contained a mixture of sodium and hydrogen, what color filter would you use to measure only the hydrogen spectrum
Answer:
Blue or Purple color filter
Explanation:
Given that Sodium has an emission spectrum with wavelength ≈ 590nm and for a wavelength of 590nm the color is yellow.
Hence To filter out the color ( yellow ) to enable the measurement of the Balmer series of hydrogen spectrum, we have to use a filter that possess the complementary color of yellow ( i.e. purple(RYB color model) or blue (RGB additive color model )
therefore color filter to be used = Blue or Purple
Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
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What particules make up the nucleus
Answer:
nucleus is a collection of particles called protons,which are positively charged..and neutrons which are electrically neutral..electrons which are negatively charged..and neutrons are in turn made up of particles called quarks ..
Explanation:
hope this helps u ...
Answer:
The Nucleus is made up of protons and neutrons.
Karl-Anthony is trying to plate gold onto his silver ring. He constructs an electrolytic cell using his ring as one of the electrodes. He runs this cell for 94.3 minutes at 205.3 mA. How many moles of electrons were transferred in this process
Answer:
0.012 moles of electrons were transferred
Explanation:
We can find the number of electrons transferred from the time in seconds and the current in Amperes using the equation:
n = I*t / F
Where n are moles of electrons transferred
I is current in Amperes = 0.2053A
t is time in seconds:
94.3min*(60s/1min) = 5658s
F is faraday constant 96485A*s/mol
Replacing:
n = 0.2053A*5658s / 96485A*s/mol
n = 0.012 moles of electrons were transferred
Moles are the mass per unit molar mass of compound. The number of moles of electrons that are transferred is 0.012 moles.
What is an electrolytic cell?An electrolytic cell is a type of electrochemical cell that uses electrical energy from external sources to conduct the chemical energy in a cell.
The moles transferred in electrolytic cells are measured as:
[tex]\rm n = \rm \dfrac{I\times t}{F}[/tex]
Given,
Current (I) in amperes = 0.2053 A
Time (t) in seconds = 5658 sec
Faraday constant (F) = 96485 A -s/mol
Substituting values in the equation above moles (n) can be calculated as:
[tex]\begin{aligned} \rm n &= \dfrac{ 0.2053 \times 5658 }{96485}\\\\&= 0.012 \;\rm mol\end{aligned}[/tex]
Therefore, 0.012 moles are transferred in this process.
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If 650. grams of C6H12O6 (FW 180.16) reacts with 650. grams of O2 (FW 32.00) in cellular respiration: C6H12O6 6 O2 --> 6 CO2 6 H2O Which is the limiting reactant
Answer:
O2 is limiting reactant
Explanation:
To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:
Moles C6H12O6:
650g * (1mol/180.16g) = 3.608 moles C6H12O6
Moles O2:
650g * (1mol/32g) = 20.31 moles O2
Now, for a complete reaction of 3.608 moles of C6H12O6 are required:
3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2
As there are just 20.31 moles of O2,
O2 is limiting reactantplease see attachment
Answer:
I'll see it
Explanation:
A sample of helium gas at a pressure of 0.778 atm and a temperature of 20.2 °C, occupies a volume of 574 mL. If the gas is allowed to expand at constant temperature until its pressure is 0.373 atm, the volume of the gas sample will be mL.
Answer:
I don't know ❤️
……….is strong due to the ……………..between positive ions and negative delocalized electrons
Answer:
atom &bond
Explanation:
atom is strong due to the bond
What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?
Chemistry 11 Solutions
978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85
Amount in moles, n, of the NaCl(s):
NaCl
2.5 g
m
n
M
58.44 g
2
4.2778 10 m l
ol
o
/m
u
Molar concentration, c, of the NaCl(aq):
–2 4.2778 × 10 mol
0.100
0.42778 mol/L
0.43 mol
L
/L
n
c
V
The molar concentration of the saline solution is 0.43 mol/L.
Check Your Solution
The units are correct and the answer correctly shows two significant digits. The
dilution of the original concentrated solution is correct and the change to mol/L
seems reasonable.
Section 8.4 Preparing Solutions in the Laboratory
Solutions for Practice Problems
Student Edition page 386
51. Practice Problem (page 386)
Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,
(NH4)2SO4(aq).
What volume of the stock solution do you need to use to prepare each of the
following solutions?
a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)
b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)
c. 250 mL of 0.300 mol/L NH4
+
(aq)
What Is Required?
You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution
needed to prepare each given dilute solution.
The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.
What is dilution?Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.
Given,
Initial volume = V₁
Initial molar concentration (M₁) = 1.50 mol/L
Final volume (V₂) = 125 mL = 0.125 L
Final molar concentration (M₂)= 0.60 mol/L
The dilution is calculated as:
M₁V₁ = M₂V₂
V₁ = M₂V₂ ÷ M₁
Substituting the values in the above formula as
V₁ = M₂V₂ ÷ M₁
V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L
V₁ = 0.05 L
= 50 mL
Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.
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Because the double bond in an alkene is rigid, alkenes can exist as geometric isomers. To clarify geometric isomers, IUPAC uses cis- and trans- as part of a compound name. If the substituents around the double bond are on the same side of the double bond, this is called
cis, cis.
cis.
cis, trans.
trans.
Answer:
cis
Explanation:
Cis isomers are formed when the substituents on the carbons of the double bond are on the same side of the double bond, forming a U. Trans isomers have substituents on opposite sides of the double bond, forming a sideways Z.
The rate constant of an SN1 reaction depends on the nucleophile b. The rate constant of an SN2 reaction does not depend on the nucleophile c. SN1 reactions proceed via carbocation intermediates d. The SN2 mechanism does not involve an intermediate
Answer:
The rate constant of an SN1 reaction depends on the nucleophile
The rate constant of an SN2 reaction does not depend on the nucleophile
Explanation:
Let us recall that in an SN1 reaction, the rate determining step involves only the alkyl halide substrate and not the nucleophile. Hence;
Rate = k[RX]
Therefore;
k= Rate/[RX]
For an SN2 reaction, the rate determining step involves both the nucleophile and the alkyl halide substrate.
Hence;
Rate = k[Nu-] [RX]
k= Rate/[Nu-] [RX]
Note that;
[Nu-] = concentration of the nucleophile
[RX] =concentration of alkyl halide substrate
k= rate constant
We can see from the above derivations that;
1) The rate constant of an SN1 reaction does not depend on the nucleophile
2) The rate constant of an SN2 reaction depends on the nucleophile
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
lution: What is the molarity of 245 g of H, SO4 dissolved in 1.00 L of solution?
Answer:
Cm = n/V
n(H2SO4) = 245/98 = 2.5 mol
Cm(H2SO4) = 2.5/1 = 2.5 M
Explanation:
Complete (predict the products, write correct formulas) and balance the following reaction:
Gaseous hydrochloric acid reacts with solid calcium hydroxide. Include states of matter and use subscripts as needed to receive full credit for a correct answer.
Explanation:
since hydrochloric acid is an acid and calcium hydroxide is a base, we know that this is an acid base reaction. The ions will then dissociatiate and bond with one another.
Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is Choose... oC and the boiling point of toluene is Choose... oC. Therefore, the liquid collected first should be Choose... .
Answer:
81°C
111°C
cyclohexane
Explanation:
Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.
Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.
What mass of precipitate (in g) is formed when 250.0 mL of 0.150 M CuCl₂ is mixed with excess KOH in the following chemical reaction?
CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq)
Answer:
3.6487g
CuCl2 moles reacted = (0.15×250)/1000
according to balanced chemical equation
precipitated Cu(OH)2 moles = Reacted CuCl2 moles
molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5
mass of precipitate = (97.5 × 0.15×250)/1000
= 3.648g
9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14
