If the half-life of a radioisotope is 20,000 years, then a sample in which three-quarters of that radioisotope has decayed is ___________ years old
An experiment looking at structures smaller than a cell would most likely employ a _______.
a.
dissecting microscope
b.
transmission electron microscope
c.
scanning electron microscope
d.
compound light microscope
what quantity and unit are measured using a balance
The heat of reaction for the combustion of c2h4 is -1411 kj/mol. how many grams of c2h4 must burn
Final answer:
To release -1411 kJ of energy, 28.054 grams of [tex]C_2H_4[/tex] needs to burn since the heat of reaction is directly proportional to the amount of the substance.
Explanation:
The question asks for the amount of [tex]C_2H_4[/tex] (ethylene) that needs to be burned given the heat of reaction for its combustion is -1411 kJ/mol. First, it's essential to understand that the heat of reaction indicates how much energy is released (in this case, -1411 kJ) when 1 mole of a substance undergoes a chemical reaction. To find out how many grams of [tex]C_2H_4[/tex] must burn to release a specific amount of energy, one has to relate this energy value to the molar mass of [tex]C_2H_4[/tex].
The molar mass of [tex]C_2H_4[/tex] is 28.054 g/mol, which means 28.054 grams of [tex]C_2H_4[/tex] constitute one mole. Given the energy released is directly proportional to the amount of substance, to release -1411 kJ, one mole or 28.054 grams of [tex]C_2H_4[/tex] needs to burn. If a different amount of energy is desired, the amount of [tex]C_2H_4[/tex] needed can be calculated proportionally based on the given heat of reaction.
The ka value for acetic acid, ch3cooh(aq), is 1.8× 10–5. calculate the ph of a 1.60 m acetic acid solution.
To calculate the pH of a 1.60 M acetic acid solution, use the ionization constant (Ka) of acetic acid, set up an ICE table, and solve for x to find the concentration of the hydronium ion (H3O+). Then, calculate the pH using the equation pH = -log[H3O+]. The pH of the acetic acid solution is approximately 1.5873.
Explanation:To calculate the pH of a 1.60 M acetic acid solution, we can use the ionization constant (Ka) of acetic acid and the relationship between the concentration of the acid, its conjugate base, and the pH. The Ka value for acetic acid is 1.8×10-5. Since acetic acid is a weak acid, we can assume that the concentration of the hydronium ion (H3O+) is equal to the concentration of the acid that has ionized.
Using the equation Ka = [H3O+][C2H3O2]/[HC2H3O2], we can set up an ICE table to calculate the concentration of the hydronium ion:
Initial concentration: [H3O+] = unknown, [C2H3O2] = 0, [HC2H3O2] = 1.60 MChange in concentration: [H3O+] = -x, [C2H3O2] = x, [HC2H3O2] = -xEquilibrium concentration: [H3O+] = 1.60 - x, [C2H3O2] = x, [HC2H3O2] = 1.60 - xSubstituting these values into the Ka expression, we get 1.8×10-5 = (1.60 - x)(x)/(1.60 - x). Solving for x gives us x ≈ 0.0127 M. Since [H3O+] = 1.60 - x, the pH of the solution is approximately 1.60 - 0.0127 = 1.5873.
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The pH of a 1.60 M acetic acid solution can be calculated using the equilibrium constant expression and initial concentration. The pH is equal to -log[H3O+]. The pH of this particular solution is 2.37.
Explanation:The pH of a 1.60 M acetic acid solution can be calculated using the equation:
pH = -log[H3O+]
First, we need to find the concentration of the hydronium ion, [H3O+]. To do this, we use the equilibrium constant expression for acetic acid, which is Ka = [H3O+][CH3CO2-] / [CH3COOH]. With a given Ka value of 1.8 × 10-5 and a concentration of acetic acid of 1.60 M, we can plug in these values to find the concentration of [H3O+], which is then used to calculate the pH.
Therefore, the pH of a 1.60 M acetic acid solution is 2.37.
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What is true of a basic solution at room temperature? it has a ph value below 7. it has a greater concentration of hydroxide compared to hydronium ions. it has a distinct sour taste but an odorless gas. it can be used as a conductor in car batteries?
A considerable amount of heat is required for the decomposition of aluminum oxide. 2 al2o3(s) â 4 al(s) + 3 o2(g) δh = 3352 kj (a) what is the heat change for the formation of 1 mol of aluminum oxide from its elements?
The heat change for the formation of 1 mol of aluminum oxide from its elements is 1676 kJ.
Explanation:The heat change for the formation of 1 mol of aluminum oxide from its elements can be calculated using the information provided in the question. We know from the given thermochemical equation that the decomposition of 2 mol of aluminum oxide releases 3352 kJ of heat. Therefore, we can set up a proportion to find the heat change for the formation of 1 mol of aluminum oxide:
3352 kJ / 2 mol = x kJ / 1 mol
Solving for x gives us the heat change for the formation of 1 mol of aluminum oxide, which is 1676 kJ.
Boats typically have several metal fittings where ropes can be fastened. What is the name for these metal fittings?
The metal fittings on boats where ropes can be fastened are called cleats. Cleats are typically made of metal or plastic and are used to secure ropes by wrapping them around the fitting. Cam cleats are a specific type that use cams to grip and release the rope easily.
The metal fittings on boats where ropes can be fastened are called cleats. Cleats come in various shapes and materials, including metal and plastic, and are designed to secure ropes by wrapping them around the fitting. A specific type of cleat often used in sailing is the cam cleat, which uses cams to grip the rope when pulled upward and allows easy release when pulled downward.
In addition to cleats, there are other fittings like bitts and bollards that also serve the purpose of securing ropes on boats. However, cleats are the most commonly used and recognized for this function.
How many moles of H2O can be formed when 4.5 moles of NH3 react with 3.2 moles of O2?
4NH3 + 5O2 yields 4NO + 6H2O
A)3.8 mol H2O
B) 4.8 mol H2O
C) 6.4 mol H2O
D) 6.8 mol H2O
For the weak acid ch3cooh (acetic acid) that is titrated with a strong base (naoh), what species (ions/molecules) are present in the solution at the stoichiometric point?
At the stoichiometric point of the titration of acetic acid with a strong base, the species present in the solution are acetate ions (CH3COO-) and water (H2O).
Explanation:During the titration of acetic acid (CH3COOH) with a strong base (NaOH), at the stoichiometric point all of the acetic acid will have reacted with the sodium hydroxide. This means that the species present in the solution at the stoichiometric point are the acetate ions (CH3COO-) and water (H2O). The balanced chemical equation for the reaction is:
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
Earth's climate has been stable throughout its history. True False
Adding a base tends to _____ of a solution. (etext concept 3.3)
Adding a base to a solution reduces its acidity by releasing hydroxyl ions or absorbing H* already present in the solution. The relative strength of a base is indicated by its base-ionization constant, and buffered solutions experience slight changes in pH when acid or base is added.
Explanation:Adding a base to a solution tends to reduce the acidity of the solution. A base is a substance that releases hydroxyl ions (OH) in solution, or one that accepts H* already present in solution. These hydroxyl ions or other basic substances combine with H* to form a water molecule, thereby removing H* and reducing the solution's acidity. This process is reflected in the base-ionization constant (K), which reflects the relative strength of a base in solution. A stronger base has a larger ionization constant than does a weaker base, resulting in a larger reduction in solution acidity. In addition, solutions that contain appreciable amounts of a weak conjugate acid-base pair are known as buffers. These buffered solutions generally experience only slight changes in pH when small amounts of acid or base are added due to their buffer capacity.
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Explain how the law of conservation of mass necessitates our balancing all chemical equations.
1. If 100.0 grams of ethylene glycol are dissolved in 900.0 grams of water, what is the freezing temperature of the solution formed? Follow these steps: Show all of your calculations. Calculate the molar mass of ethylene glycol: 62.07 g/mol Calculate the number of moles of ethylene glycol in the solution: Calculate the molality of ethylene glycol: Calculate the freezing point depression using the Kf from the Chemistry B Information Sheet and the molality that you calculated: Calculate the freezing point of the solution. 2. Think about the result. Think about the typical mid-winter temperature we experience in Michigan. Is this concentration of ethylene glycol high enough to use in a car radiator in the winter? Why or why not? please hurry
1. How many milliliters of a 0.184 M NaNO3 solution contain 0.113 moles of NaNO3?ans:614 mL
Explanation:
The given data is as follows.
Molarity of solution = 0.184 M
Volume of solution = ?
number of moles = 0.113 mol
As molarity is the number of moles present in liter of solvent.
Mathematically, Molarity = [tex]\frac{\text{No. of moles}}{\text{Volume}}[/tex]
Hence, calculate the volume of given solution as follows.
Molarity = [tex]\frac{\text{No. of moles}}{\text{Volume}}[/tex]
0.184 M = [tex]\frac{0.113 mol}{volume}[/tex]
volume = 0.614 L
As 1 L = 1000 mililiter. Hence convert 0.614 L into ml as follows.
[tex]0.614 L \times \frac{1000 ml}{1 L}[/tex]
= 614 ml
Thus, we can conclude that the volume of given solution is 614 ml.
When 7.0 mol Al react with 8.5 mol HCl, what is the limiting reactant and how many moles of AlCl3 can be formed?
2 Al + 6 HCl yields 2 AlCl3 + 3 H2
Al is the limiting reactant; 7.0 mol AlCl3 can be formed
HCl is the limiting reactant; 2.8 mol AlCl3 can be formed
Al is the limiting reactant; 3.5 mol AlCl3 can be formed
HCl is the limiting reactant; 8.5 mol AlCl3 can be formed
B. HCl is the limiting reactant; 2.8 mol AlCl3 can be formed
When you hear the word decahydrate, what number should be written in front of the formula for water?
How many moles of nacl are required to prepare 0.80 l of 6.4 m nacl? 0.13 mol nacl 5.1 mol nacl 7.2 mol nacl 8.0 mol nacl?
Answer:
5.12 moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl.
Explanation:
Molarity (M) is a concentration measure that indicates the number of moles of solute that are dissolved in a given volume.
The Molarity is then determined by:
[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units[tex]\frac{moles}{liter}[/tex].
6.4 M NaCl indicates that 1 liter of solution there are 6.4 moles of NaCl.
You can apply a rule of three as follows: if in 1 liter of solution there are 6.4 moles of NaCl, in 0.8 L how many moles are there?
[tex]moles=\frac{0.8 L*6.4 moles}{1 L}[/tex]
moles=5.12
5.12 moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl.
How many atoms are in 15.0 moles of C2H6O
Answer: [tex]813.1\times 10^{23}[/tex] atoms.
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
1 molecule of [tex]C_2H_6O[/tex] contains = 9 atoms
1 mole of [tex]C_2H_6O[/tex] contains = [tex]9 \times 6.023\times 10^{23}=54.21\times 10^{23}[/tex] atoms
Thus 15 moles [tex]C_2H_6O[/tex] contains =[tex]\frac{54.21\times 10^{23}}{1}\times 15=813.1\times 10^{23}[/tex] atoms.
Thus the answer is [tex]813.1\times 10^{23}[/tex] atoms.
[Gradpoint-Question AK-4] Which equation represents a combustion reaction?
A: Ca + 2H Cl -----> CaCl₂ + H₂
B: Pb(NO₃)₂ +2HCl -----> PbCl₂ + 2HNO₃
C: 2SO₂ + O₂ ------> 2SO₃
D: 2C₂H₆ + 7O₂ -----> 4CO₂ + 6H₂O
This question is worth 40 points. So please answer carefully. I'm bad at Chemical equations.
The equation represents a combustion reaction is -
D: 2C₂H₆ + 7O₂ -----> 4CO₂ + 6H₂OA reaction involves a substance reacting with oxygen and releasing energy in form of light or heat, such a chemical reaction is called a combustion reaction.
Burning coal, methane gas, and sparklers are all common examples of combustion reactions2C₂H₆ + 7O₂ -----> 4CO₂ + 6H₂O
Here ethane reacts with oxygen and releases carbon dioxide and water with a high amount of energy. Pb(NO₃)₂ +2HCl -----> PbCl₂ + 2HNO₃ is a double replacement, acid-base reaction. It is also called a neutralization reaction.2SO₂ + O₂ ------> 2SO₃ is a combination reaction that is an example of an oxidation-reduction reaction. Ca + 2H Cl -----> CaCl₂ + H₂ is a single replacement reaction that is also an example of oxidation-reduction reactionThus, the equation represents a combustion reaction is -
D: 2C₂H₆ + 7O₂ -----> 4CO₂ + 6H₂OLearn more:
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How many total carbon atoms are found in a molecule of 3-ethyl-2-pentene? 5 6 7 8?
Answer:
The correct answer is 7.
Explanation:
Given : 3-ethyl-2-pentene (Alkene)
The longest chain in the molecule is of 5 carbon atom. And on second carbon ethyl group is present.In ethyl group there re 2 carbon atoms. So, the total carbon atom in 3-ethyl-2-pentene are 7 carbon atoms.
[tex]CH_2=C(C_2H_5)-CH_2-CH_2-CH_3[/tex]
Determine the specific heat capacity of an alloy that requires 59.3 kj to raise the temperature of 150.0 g alloy from 298 k to 398 k.
To determine the specific heat capacity of an alloy, use the formula Q = mcΔT and rearrange for c. Substitute the given values and calculate the answer.
Explanation:The specific heat capacity of an alloy can be calculated using the formula:
Q = mcΔT
Where Q is the heat energy, m is the mass of the alloy, c is the specific heat capacity, and ΔT is the change in temperature. Rearranging the formula to solve for c, we have:
c = Q / (mΔT)
Substituting the values given in the question, we get:
c = 59.3 kJ / (150.0 g × (398 K - 298 K))
After performing the calculations, the specific heat capacity of the alloy is obtained.
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The specific heat capacity of an alloy that requires 59.3 KJ to raise the temperature of 150.0 g alloy from 298 K to 398 K is 3.95 J/g·K.
To determine the specific heat capacity (c) of an alloy, we can use the formula:
q = mcΔT
where:
q represents the heat absorbed (in joules)m is the mass (in grams)ΔT is the change in temperature (in Kelvin or degrees Celsius)c is the specific heat capacity (in J/g·K)Given the information:
q = 59.3 kJ = 59,300 Jm = 150.0 gΔT = 398 K - 298 K = 100 KPlugging these values into the formula:
59,300 J = 150.0 g × c × 100 K
Solving for c:
c = (59,300 J) / (150.0 g × 100 K) = 3.95 J/g·K
Therefore, the specific heat capacity of the alloy is 3.95 J/g·K.
A saturated straight-chain hydrocarbon with two carbons is ____.
Answer:
A saturated sraight-chain hydrocarbon with two carbons is ethane.
Explanation:
Saturated hydrocarbons are called as alkanes with general formula [tex]C_{n}H_{2n+2}[/tex] where n=1,2,3,4......
Now, as it contains two carbon atoms therefore name of this compound should contain a prefix "eth". Also a suffix "ane" will come due to the fact that it is a saturated hydrocarbon.
So, molecular formula of this compound is [tex]C_{2}H_{6}[/tex]
Structure of this straight chain compound has been attached below.
Which of the following is part of the lithosphere?
An unsaturated solution is one that ________. an unsaturated solution is one that ________. contains the maximum concentration of solute possible, and is in equilibrium with undissolved solute has no double bonds contains no solute has a concentration lower than the solubility contains more dissolved solute than the solubility allows
Sodium hydrogen carbonate nahco3 , also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid hcl , which the stomach secretes to help digest food. drinking a glass of water containing dissolved nahco3 neutralizes excess hcl through this reaction: hcl (aq) nahco3 (aq) â nacl (aq) h2o (l) co2 (g) the co2 gas produced is what makes you burp after drinking the solution. suppose the fluid in the stomach of a man suffering from indigestion can be considered to be 250.ml of a 0.076 m hcl solution. what mass of nahco3 would he need to ingest to neutralize this much hcl
Answer:
1.596 g
Explanation:
The neutralization reaction is:
NaHCO₃ + HCl → NaCl + H₂O + CO₂
The number of moles of the acid in stomach is the volume (250.0 mL = 0.250 L) multiplied by the molar concentration (0.076 M):
n = 0.250 * 0.076 = 0.019 mol of HCl
By the stoichiometry of the reaction, 1 mol of NaHCO₃ is needed to neutralize 1 mol of HCl, so it'll be necessary 0.019 mol of sodium bicarbonate.
The molar mass of NaHCO₃ is 84 g/mol, thus the mass of it is:
m = number of moles * molar mass
m = 0.019 * 84
m = 1.596 g
A solution with a hydrogen ion concentration of 3.25 × 10-2 m is ________ and has a hydroxide concentration of _______
To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:
pH = - log [H+] where H+ must be in Molar
We are given that H+ = 3.25 × 10-2 M
Therefore the pH is:
pH = - log [3.25 × 10-2]
pH = 1.488
Since pH is way below 7, therefore the solution is acidic.
To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,
[H+]*[OH-] = 10^-14
[OH-] = 10^-14 / [H+]
[OH-] = 10^-14 / 3.25 × 10-2
[OH-] = 3.08 × 10-13 M
Answers:
Acidic
[OH-] = 3.08 × 10-13 M
The hydrogen ion concentration is 3.25 × 10-2 M, indicating an acidic solution with a pH of 1.49. The hydroxide concentration is 3.08 × 10-13 M.
Explanation:In this question, the hydrogen ion concentration is given as 3.25 × 10-2 M. This concentration indicates that the solution is acidic, with a pH value equal to -log[H3O+]. Therefore, pH = -log(3.25 × 10-2) = 1.49.
The hydroxide concentration can be calculated using the formula: [OH-] = Kw/[H3O+]. At 25°C, the value of Kw (the ion-product constant for water) is 1.0 × 10-14. Therefore, [OH-] = 1.0 × 10-14 / 3.25 × 10-2 = 3.08 × 10-13 M.
A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a mass of 0.333 g. the limiting reactant in the salt mixture is k2c2o4⋅h2o. cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) à cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l) starting material (sm) product molar mass (mm) g/mol: cacl2⋅2h2o = 147.02 k2c2o4⋅h2o = 184.24 cac2o4 = 128.10 determine mass of k2c2o4⋅h2o(aq) in salt mixture in grams. answer to 3 places after the decimal and include unit, g
Answer: The mass of [tex]K_2C_2O_4.H_2O[/tex] in the salt mixture is 0.424 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For [tex]CaC_2O_4.H_2O[/tex] :Given mass of [tex]CaC_2O_4.H_2O[/tex] = 0.333 g
Molar mass of [tex]CaC_2O_4.H_2O[/tex] = 146.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of}CaC_2O_4.H_2O=\frac{0.333g}{146.12g/mol}=0.0023mol[/tex]
The given chemical equation follows:
[tex]CaCl_2.2H_2O(aq.)+K_2C_2O_4.H_2O(aq.)\rightarrow CaC_2O_4.H_2O(s)+2KCl(aq.)+2H_2O(l)[/tex]
By Stoichiometry of the reaction:
1 mole of [tex]CaC_2O_4.H_2O[/tex] is produced by 1 mole of [tex]K_2C_2O_4.H_2O[/tex]
So, 0.0023 moles of [tex]CaC_2O_4.H_2O[/tex] will be produced by = [tex]\frac{1}{1}\times 0.0023=0.0023mol[/tex] of [tex]K_2C_2O_4.H_2O[/tex]
Now, calculating the mass of [tex]K_2C_2O_4.H_2O[/tex] by using equation 1, we get:
Molar mass of [tex]K_2C_2O_4.H_2O[/tex] = 184.24 g/mol
Moles of [tex]K_2C_2O_4.H_2O[/tex] = 0.0023 moles
Putting values in equation 1, we get:
[tex]0.0023mol=\frac{\text{Mass of }K_2C_2O_4.H_2O}{184.24g/mol}\\\\\text{Mass of }K_2C_2O_4.H_2O=(0.0023mol\times 184.24g/mol)=0.424g[/tex]
Hence, the mass of [tex]K_2C_2O_4.H_2O[/tex] in the salt mixture is 0.424 grams.
Final answer:
The mass of K2C2O4·H2O in the salt mixture is calculated using the molar mass and the moles of the precipitate formed in the reaction, resulting in a mass of 0.479 g.
Explanation:
To determine the mass of K2C2O4·H2O in the salt mixture, we start by identifying that the precipitate formed, CaC2O4·H2O, has a molar mass of 128.10 g/mol and a measured mass of 0.333 g. Using this information, the number of moles of CaC2O4·H2O can be calculated as moles = mass / molar mass = 0.333 g / 128.10 g/mol.
This results in 2.60 x 10-3 moles of CaC2O4·H2O. Since the reaction shows that K2C2O4·H2O and CaCl2·2H2O react in a 1:1 mole ratio to form the precipitate, the same number of moles is used for K2C2O4·H2O. The mass of K2C2O4·H2O can be found by multiplying the number of moles by its molar mass (2.60 x 10-3 moles x 184.24 g/mol).
The resulting mass of K2C2O4·H2O in the salt mixture is 0.479 g, to three decimal places.
What is the molar concentration of potassium ions in a 0.250 m k2so4 solution?
Answer : The molar concentration of [tex]K^+[/tex] ion is, 0.5 M
Explanation :
The dissociation reaction of [tex]K_2SO_4[/tex] is,
[tex]K_2SO_4(aq)\rightarrow 2K^+(aq)+SO_4^{2-}(aq)[/tex]
By the stoichiometry we can say that, 1 mole of [tex]K_2SO_4[/tex] dissociates into 2 mole of [tex]K^+[/tex] ions and 1 mole of [tex]SO_4^{2-}[/tex] ions.
As we are given the concentration of [tex]K_2SO_4[/tex] is, 0.250 M.
So, the molar concentration of [tex]K^+[/tex] ion = [tex]2\times 0.250=0.5M[/tex]
The molar concentration of [tex]SO_4^{2-}[/tex] ion = 0.250 M
Therefore, the molar concentration of [tex]K^+[/tex] ion is, 0.5 M
Given an initial cyclopropane concentration of 0.00560 m, calculate the concentration of cyclopropane that remains after 1.50 hours.
We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:
A = Ao e^(- k t)
Where,
A = amount remaining at
time t = unknown (what to solve for)
Ao = amount at time zero = 0.00560
M
k = rate constant
t = time = 1.50 hours or 5400 s
The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:
k = 5.29× 10^–4 s–1 (plug in the correct k value)
Plugging in the values in the 1st equation:
A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )
A = 3.218 × 10^–4 M (simplify as necessary)
To calculate the concentration of cyclopropane that remains after 1.50 hours, we need to use the first-order integrated rate law equation. Without the rate constant, we cannot calculate the exact concentration, but we can determine that it will be less than 0.200 mol/L.
To calculate the concentration of cyclopropane that remains after 1.50 hours, we need to use the first-order integrated rate law. The rate constant for the reaction is not given, so we cannot calculate it. However, we can use the given information that at 10.0 minutes (0.1667 hours) the concentration of cyclopropane is 0.200 mol/L and find the concentration at 1.50 hours using the integrated rate law equation:
[tex][C] = [C]0 * e^(-kt)[/tex]
where [C] is the concentration at time t, [C]0 is the initial concentration, k is the rate constant, and t is the time.
Given [C]0 = 0.200 mol/L, t = 1.50 hours, and [C] = ?
Let's solve the integrated rate law equation:
Plug in the known values:
[tex][C] = 0.200 * e^(-k*1.50)[/tex]
Since the rate constant is not given, we cannot calculate the exact concentration. However, we can still make a general qualitative statement that the concentration will be less than 0.200 mol/L.
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