Answer:
Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.
Explanation:
There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm. The atomic radius of aluminium atom is 143 ppm. The atomic radius of silicon atom is 111 ppm. The atomic radius of phosphorus atom is 98 ppm. The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.
One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Answer:
6.5 mg/L.
Explanation:
Step one: write out and Balance the chemical reaction in the Question above:
NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.
Step two: Calculate or determine the number of moles of AgCl.
So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:
Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.
Step three: Calculate or determine the number of moles of NiCl2.
Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.
Step four: detemine the mass of NiCl2.
Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.
Step five: finally, determine the concentration of NiCl2.
1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.
Provide the name(s) for the tertiary alcohol(s) with the chemical formula C6H14O that have a 4-carbon chain. Although stereochemistry may be implied in the question, DO NOT consider stereochemistry in your name. Alcohol #1______ Alcohol #2: ______Alcohol #3______
Answer:
Explanation:
A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.
Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is
2-hydroxy-2,3-dimethylbutane
H OH H H
| | | |
H - C - C - C - C - H
| | | |
H CH₃ CH₃ H
From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain
Identify a homogeneous catalyst. Identify a homogeneous catalyst. Pd in H2 gas N2 and H2 catalyzed by Fe SO2 over vanadium (V) oxide Pt with methane H2SO4 with concentrated HCl
Answer:
H2SO4 with concentrated HCl
Which of the following pieces of information is given in a half-reaction?
O A. The number of electrons transferred in the reaction
B. The compounds that the atoms in the reaction came from
C. The state symbol of each compound in the reaction
D. The spectator ions that are involved in the reaction
Answer:
The number of electrons transferred in the reaction
Explanation:
Answer:
A
Explanation:
Which solution, if either, would create the higher osmotic pressure (compared to pure water): one prepared from 1.0 g of NaCl in 10 mL of water or 1.0 g of CsBr in 10 mL of water
Answer: NaCl would give the higher pressure
Explanation:
Osmotic pressure depends only on the number of ions.
NaCl dissociates as Na+ and Cl- ; CsBr dissociates as Cs+ and Br-
But the concentration of the solutions are different.
Concentration (morality ) of NaCl = Moles /Litre = (1 g /58.44g/mol)/0.01L
Total number of ions in NaCl solution = 2 x (1 g /58.44g/mol)/0.01L ( 1 mol NaCl gives 2 moles ions, 1 mol Na+ and 1 mol Cl-)
= 1.71×2RT
Similarly total number of ions in CsBr solution = 2 x (1 g /212.80 g/mol)/0.01L
= 0.47×2RT
Therefore osmotic pressure is higher in NaCl solution.
Given that π = n M R T, rearrange the equation to solve for V
Answer:
V= n/M
Explanation:
From;
π = nRT/V = MRT
Where;
n= number of moles
R= gas constant
T= absolute temperature
M= molar mass
V= volume of the solution
π= osmotic pressure
Thus;
nRT/V = MRT
nRT = VMRT
V= nRT/MRT
V= n/M
There are approximately 2 × 1022 molecules and atoms in each breath we take and the concentration of CO in the air is approximately 9 ppm. How many CO molecules are in each breath we take? solution
Answer:
1.8x10¹⁷ molecules of CO are in each breath we take
Explanation:
Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.
A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.
In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:
2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =
1.8x10¹⁷ molecules of CO are in each breath we take[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take
The calculation is as follows:A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.
Now CO molecules in each breath is
[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]
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What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
The surface temperature on Venus may approach 753 K. What is this temperature in degrees Celsius?
Answer:
461.85 degrees Celsius
Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–5 M 3.0 × 10–2 M 2.4 × 10–2 M None of the above.
Answer:
3.0x10⁻²M
Explanation:
Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:
Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:
ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:
1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
1.4x10⁻⁵ = [2X]² [X]
1.4x10⁻⁵ = 4X³
3.5x10⁻⁶ = X³
0.015 = X
As [Ag⁺] is 2X:
[Ag⁺] = 0.030 = 3.0x10⁻²M
The answer is:
3.0x10⁻²MIf the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?
Answer:
6.93
Explanation:
Step 1: Given data
Standard Gibbs free energy (∆G°): -5.20 kJTemperature (T): 50°CEquilibrium constant (K): ?Step 2: Convert the temperature to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 50°C + 273.15
K = 323 K
Step 3: Calculate K
We will use the following expression.
∆G° = -R × T × ln K
-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K
K = 6.93
Chlorine has an electronegativity value of 3.0. Given the electronegativity of N, O, and P (3.0, 3.5, and 2.1, respectively), which of the following molecules has nonpolar bonds?a. NCl3b. Cl2Oc. PCl3d. All of thesee. None of these
Answer:
a. NCl3
Explanation:
All the elements stated are non metals. Generally the bond between non metals is a covalent bond. However depending on how the electrons are shared, this bond can either be polar or non polar.
A non polar covalent bond is formed when the electrons are shared equally between the atoms.
A way to determine if a bond is non polar or polar covalent is by comparing the electronegativity values.
If the electronegativity difference is less than 0.5, it is regarded as a non polar covalent bond.
Going back to the question;
Between N and Cl; the electronegativity difference is 3.0 - 3.0 = 0
Between Cl and O; the electronegativity difference is 3.5 - 3.0 = 0.5
Between Cl and P; the electronegativity difference is 3.5 - 2.1 = 1.4
From these we can tell that the correct option is option A.
The absorption spectrum of argon has a line at 515 nm. What is the energy of
this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's
constant is 6.626 x 10-34 Jos.)
O A. 2.59 x 1027j
O B. 3.86 x 10-28 J
O C. 3.86 x 10-19 J
O D. 2.59 x 1018 J
Answer:
OPTION C is correct
3.86 x 10-19 J
Explanation:
Energy of the line can be calculated using below formula
E= h ν.................(1)
Where E= energy
h= plank constant= 6.626 10-34 J s
c=speed of light=3 x 108 m/s
But we know that Velocity V= = c / λ
Then substitute into equation (1) we have
E = h c / λ.............(2)
We can calculate our( hc ) in nm for unit consistency
h c =( 6.626 ×10^-34)x(3×108)
h c = (1.986 x 10-16 )
hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)
E = h c / λ = (1.986 x 10-16) / 515
E = 3.86 x 10-19 J
Therefore, the Energy is 3.86 x 10-19 J
g All of the molecules below have polar bonds but only one of the molecules is a polar molecule. Which one is a polar molecule? A) C2F2 B) C2Cl4 C) CO2 D) NF3 E) CF4
Answer: [tex]NF_3[/tex]
Explanation:
Geometrical symmetry of the molecule and the polarity of the bonds determine the polarity of the molecule.
The molecule that has zero dipole moment that means it is a geometrically symmetric molecule and the molecule which has some net dipole moment means it is a geometrically asymmetric molecule.
As the molecule is symmetric, the dipole moment will be zero as dipole moments cancel each other and the molecule will be non-polar.
As the molecule is asymmetric, the dipole moment will not be zero and the molecule will be polar.
Example: [tex]NF_3[/tex]
Thus, we can say that [tex]NF_3[/tex] is a polar molecule.
a jogger runs a mile in 8.92 minutes. 1 mi=1609m; calculate her speed in km/hr
Answer:
[tex]Speed = 3.30 \frac{km}{hr}[/tex]
Explanation:
Given
Distance = 1 mile
Time = 8.92 minutes
Required
Calculate Speed in km/hr
Speed is calculated as thus;
[tex]Speed = \frac{Distance}{Time}[/tex]
Substitute 1 mile for distance and 8.92 minutes for time
[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]
Convert Miles to Kilometres
If
[tex]1\ mile = 1609\ m[/tex];
Then
[tex]1\ mile = \frac{1609\ km}{1000}[/tex]
[tex]1\ mile = 1.609\ km[/tex] --- (1)
Convert minutes to hour
[tex]1\ minutes = 0.0167\ hour[/tex]
Multiply both sides by 8.92
[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]
[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)
By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have
[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]
[tex]Speed = 3.304309 \frac{km}{hr}[/tex]
[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated
Testbank Question 47 Consider the molecular orbital model of benzene. In the ground state how many molecular orbitals are filled with electrons?
Answer:
There are fifteen molecular orbitals in benzene filled with electrons.
Explanation:
Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;
There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.
There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons
The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.
All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.
[tex]pH = -log10[H+ ][/tex]. If [tex][H^+][/tex] is [tex]1.2 * 10^-^6[/tex], what is the [tex]pH[/tex]? (F5 to refresh page if you can't see it)
what is the molality of a solution
Answer: The number of moles of a solute per kilogram of solvent
Explanation:
A man weighs 185 lb. What is his mass in grams?
Please show work.
Thank you
Answer:
83914.52 grams
Explanation:
Given that,
Weight of a man is 185 lb
We need to find his weight in grams
For this, we must know the relation between lb and grams.
We know that,
1 lb = 453.592 grams
To find the mass of man in grams, the step is :
185 lb = (453.592 × 185) grams
= 83914.52 grams
So, the mass of a man is 83914.52 grams.
A chemical reaction that has the general formula of nA → (A)n is best classified as a ____ reaction. A. synthesis B. polymerization C. decomposition D. oxidation E. replacement
Answer:
B.
Explanation:
A chemical reaction that has the general formula of nA → (A)n is best classified as a polymerization reaction.
Answer:
B. Polymerization
Explanation:
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What volume (in mL) needs to be added to 69.6 mL of 0.0887 M MgF2 solution to make a 0.0224 M MgF2 solution
Answer:
The correct answer is 206 ml.
Explanation:
Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,
M₁V₁ = M₂V₂
Now putting the values in the formula we get,
0.0887 × 69.6 = 0.0224 M × V₂
V₂ = 0.0887 × 69.6 / 0.0224
V₂ = 275.6 ml
Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,
= 275.6 ml - 69.6 ml = 206 ml
A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of the precipitate is 2.9 g. Answer the questions below to determine the percent yield. 3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq) 1. a. Which product is the precipitate? b. How many moles of the precipitate would one expect to be produced from 0.030 moles of calcium nitrate? c. How many grams of solid do you expect to be produced? d. What is the percent yield?
Answer:
a. Ca₃(PO₄)₂.
b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced
c. 3.1g of Ca₃(PO₄)₂
d. Percent yield = 93.5%
Explanation:
a. Based on the reaction:
3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)
3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.
As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,
Ca₃(PO₄)₂b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate
0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =
0.010 moles of Ca₃(PO₄)₂ can we expect to be producedc. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;
0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =
3.1g of Ca₃(PO₄)₂d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:
[tex]\frac{2.9g}{3.1g} *100[/tex]
Percent yield = 93.5%(a) The product in solid state would be the precipitate. Hence, the precipitate would be Ca3(PO4)2
(b) From the balanced equation of the reaction: 3 moles of Ca(NO3)2 is required for 1 mole of Ca3(PO4)2
If there are just 0.030 moles of Ca(NO3)2, then"
3 moles = 1
0.030 moles = 1 x 0.030/3
= 0.01 moles of Ca3(PO4)2
In other words, 0.01 moles of the precipitate would be expected to be produced from 0.030 moles of calcium nitrate.
(c) 0.01 moles solid (Ca3(PO4)2) is expected. Mass of Ca3(PO4)2 expected:
mass = mole x molar mass
molar mass of Ca3(PO4)2 = 310.18 g/mol
mass of Ca3(PO4)2 expected to be produced = 0.01 x 310.18
= 3.1018 g
Hence, 3.1018g of solid is expected to be produced.
(d) Percentage yield = actual yield/theoretical yield x 100
= 2.9/3.1018 x 100
= 93.5%
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The standard entropy of a substance refers to its entropy at:__________.
a. absolute zero and 1 bar
b. 0°C and 1 bar
c. 25 °C and 1 bar
d. 25 °C and 0 bar
Answer:
b. 0°C and 1 bar
Explanation:
Hello,
In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.
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What subatomic particles surround the nucleus? Question 1 options: protons neutrons atoms electrons
Answer:
Electrons "surround"
Explanation:
Protons and neutrons "make up" the nucleus so they are contained "within" the nucleus meaning that electrons would "surround" the nucleus as they orbit around the nucleus
Answer:
Electrons
Explanation:
Protons and nuetrons are present inside the nucleus of an atom while the electron revolve around the nucleus in different energy levels.
According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.
A certain electrochemical cell has a cell potential of +0.34 V. Which of the following is a true statement about the electrochemical reaction?
a. The reaction is reactant favored and would be considered an electrolytic cell.
b. The reaction is reactant favored and would be considered a voltaic (galvanic) cell.
c. The reaction is product favored and would be considered an electrolytic cell.
d. The reaction is at equilibrium and is a voltaic (galvanic) cell.
e. The reaction is product favored and would be considered a voltaic (galvanic
Answer:
e. The reaction is product favored and would be considered a voltaic (galvanic) cell
Explanation:
An electrochemical cell produces electrical energy from electrochemical reactions.
A voltaic cell is a type of electrochemical cell that produces electrical energy by spontaneous electrochemical reactions. In a voltaic cell, the cell potential is always positive unlike in an electrolytic cell.
Hence, given the fact that the cell potential is positive, it is a product favoured voltaic cell.
How many grams of sodium chloride are required to make 2.00 L of a solution with a concentration of 0.100 M?
Answer:
Mass = 11.688g
Explanation:
Volume = 2.00L
Molar concentration = 0.100M
Mass = ?
These quantities are relatted by the following equation;
Conc = Number of moles / volume
Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol
Number of moles = Mass / Molar mass
Mass = Number of moles * Molar mass
Mass = 0.2mol * 58.44g/mol
Mass = 11.688g
The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
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