The rate constant of the reaction is 0.0002 s-1.
Using the formula;
lnC = lnCo - kt
Where;
C = concentration at time t
Co = initial concentration
k = rate constant
t = time taken
ln(335) = ln(502) - k(2000)
k = ln(335) - ln(502)/(-2000)
k = 5.8 - 6.2/(-2000)
k = 0.0002 s-1
For a first order reaction;
t1/2 = 0.693/k
t1/2 = 0.693/ 0.0002 s-1
t1/2 =3465 s
Learn more about first order reaction: https://brainly.com/question/1195122
What gas is evolved when sulfuric acid is added to the reaction mixture?
Suppose you wanted to monitor a pH change between 5.3 and 6.3. Which indicator would be most appropriate?
Answer:
d) bromothymol blue
Explanation:
Hello,
For that pH change between 5.3 and 6.3, we need an indicator which changes the color within that rank, in such a way, among the given options, one recalls the bromothymol blue because it turns yellow below about a pH of 6 and starts getting green above 6, therefore, one states that it is the most appropriated indicator.
Best regards.
At 50 C(celcius), Kc = 2.2 x 10^3 for the reaction 3 Fe (s) + 4 H2O (g) ? Fe3O4 (s) + 4 H2 (g) What is the value of Kp at 200oC for this reaction? A) 8.8 x 103 B) 2.2 x 103 C) 5.5 x 102 D) 3.5 x 104 E) This question cannot be answered with the information provided
At 1050 K, Kp = 0.900 for the following reaction: H2(g) + CO2(g) ? H2O(g) + CO(g) If 0.200 atm of H2 and 0.200 atm of CO2 are admitted into a rigid container and allowed to reach equilibrium, what should the equilibrium partial pressure of CO be (in atm)? [enter number with three decimal places]
In an exothermic equilibrium reaction, increasing the reaction temperature favors the formation of reactants. A) True B) False
The equilibrium-constant expression for a reaction written in one direction is the reciprocal of the one for the reaction written for the reverse direction. A) True B) False
The value of Kp at 200°C for the given reaction is 2.2 x 10^3. The equilibrium partial pressure of CO in the reaction H2(g) + CO2(g) ? H2O(g) + CO(g) is 0.200 atm. Increasing the temperature in an exothermic equilibrium reaction favors the formation of products, not reactants. The equilibrium-constant expressions for a reaction and its reverse are related by the reciprocal of each other.
Explanation:To determine the value of Kp at 200°C for the reaction 3 Fe (s) + 4 H2O (g) ? Fe3O4 (s) + 4 H2 (g), we can use the relationship between Kc and Kp. Since the reaction involves gases, we can use the equation Kp = Kc(RT)Δn, where Δn is the change in the number of moles of gas. In this case, there is no change in the number of moles of gas, so Δn = 0. Therefore, the value of Kp is equal to the value of Kc, which is given as 2.2 x 10^3 at 50°C. Thus, the value of Kp at 200°C is also 2.2 x 10^3.
To calculate the equilibrium partial pressure of CO in the reaction H2(g) + CO2(g) ? H2O(g) + CO(g), first, we need to determine the initial partial pressures of H2 and CO2. Both gases have a partial pressure of 0.200 atm when admitted into the rigid container. Since the reaction is at equilibrium, the equilibrium partial pressure of CO is the same as the initial partial pressure of CO2, which is 0.200 atm.
The statement that increasing the temperature favors the formation of reactants in an exothermic equilibrium reaction is False. In an exothermic reaction, the formation of products is favored with an increase in temperature. The reaction releases heat, so increasing the temperature shifts the equilibrium position towards the products.
The statement that the equilibrium-constant expression for a reaction written in one direction is the reciprocal of the one for the reaction written for the reverse direction is False. The equilibrium-constant expression for a reaction and its reverse are related by the reciprocal of each other, but they are not exactly the same.
Learn more about Equilibrium Reactions here:https://brainly.com/question/35821841
#SPJ2
What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M = 148.3)?
Answer:
[tex]M=1.02M[/tex]
Explanation:
Hello,
In this case, we compute the concentration as the molarity of the nitrate ion as the ratio between the nitrate ion moles and the volume of the solution as shown below:
[tex]M_{NO_3^-}=\frac{n_{NO_3^-}}{V_{solution}}[/tex]
Now, the nitrate ion moles are computed considering that in mole of magnesium nitrate, there are two moles of the nitrate ion:
[tex]n_{NO_3^-}=32.0gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.3gMg(NO_3)_2}*\frac{2molNO_3^-}{1molMg(NO_3)_2}\\n_{NO_3^-}=0.432molNO_3^-[/tex]
Finally, the requested molarity turns out into:
[tex]M_{NO_3^-}=\frac{0.432molNO_3^-}{0.425L}\\M=1.02M[/tex]
Best regards.
Final answer:
To calculate the nitrate ion concentration, convert the mass of magnesium nitrate to moles, convert volume to liters, determine the molarity of magnesium nitrate, and then multiply by 2, since each formula unit yields two nitrate ions. The concentration of nitrate ion is 1.0146 M.
Explanation:
The student is asking how to calculate the concentration of nitrate ion in a solution with a known volume and mass of magnesium nitrate. We can start by finding the molarity of magnesium nitrate and then use stoichiometry to determine the concentration of nitrate ions.
First, convert the mass of Mg(NO3)2 to moles:
Moles of Mg(NO3)2 = mass (g) / molar mass (g/mol)
32.0g / 148.3g/mol = 0.2156 mol
Next, convert the volume from mL to L:
425 mL * (1L / 1000 mL) = 0.425 L
Now calculate the molarity of the Mg(NO3)2 solution:
Molarity (M) = moles / volume (L)
0.2156 mol / 0.425 L = 0.5073 M
Each unit of Mg(NO3)2 gives two nitrate ions, so the concentration of nitrate ion is:
Concentration of NO3- = 2 × molarity of Mg(NO3)2
2 × 0.5073 M = 1.0146 M
Therefore, the concentration of nitrate ion in the solution is 1.0146 M.
A 12.0 L gas cylinder is filled with 8.00 moles of gas. The tank is stored at 35°C. What is the pressure in the tank?
List the types of intermolecular forces that exist between molecules (or basic units) in each of the following species: (a) benzene (C6H6), (b) CH3Cl,
(c) PF3, (d) NaCl, (e) CS2.
Answer:
Answer has been given below
Explanation:
Every molecules contain electrons. So, London dispersion force exists in all molecules.For polar aprotic molecules, additional dipole-dipole force exists between molecules.For ionic compounds, additional ionic interaction force exist between ions.For non polar molecules, only London dispersion force exists between molecules.Benzene is a non-polar molecule. So London dispersion force exists between benzene molecules.[tex]CH_{3}Cl[/tex] is a polar molecule. So, dipole-dipole force and London dispersion force exists between [tex]CH_{3}Cl[/tex] molecules.[tex]PF_{3}[/tex] is a polar molecule. So, dipole-dipole force and London dispersion force exists between [tex]PF_{3}[/tex] molecules.NaCl is an ionic compound. So London dispersion force and ionic interaction force exists between NaCl molecules.[tex]CS_{2}[/tex] is a nonpolar molecule. So, London dispersion force exists between [tex]CS_{2}[/tex] molecules.a sample of gas with a mass of 1.70 g occupies a volume of 961 ml at a pressure of 728 torr and a temperature of 21°C. what is the molar mass of the gas?
You have lost nearly 6 inches of top soil on your 200 acre field in the last 10 years. The problem seems to be getting even worse as drought has taken hold of your area. What are some possible solutions that you would consider to reduce your soil erosion problem?
A container of hydrogen gas has a volume of 1.46 liters, a pressure of 2.18 atm, and a temperature of 185 Kelvin. How many moles of gas are in the container?
Assume ideal gas behavior.
The answer is ______ mol.
How many carbon atoms would be in the compound named chlorobenzene
What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?
A.)The equilibrium constant, Kp, for the following reaction is 0.110 at 298 K:
NH4HS(s) NH3(g) + H2S(g)
Calculate the equilibrium partial pressure of H2S when 0.371 moles of NH4HS(s) is introduced into a 1.00 L vessel at 298 K.
PH2S = ?? atm
B.) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K:
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.47 atm at 500 K .
PPCl5 = ?? atm
PPCl3 = ?? atm
PCl2 = ?? atm
Final answer:
To calculate the equilibrium partial pressure of H2S, use the equilibrium constant expression and rearrange to solve for PH2S.
Explanation:
To calculate the equilibrium partial pressure of H2S, we need to use the equilibrium constant, Kp. The equilibrium constant expression for the given reaction is:
Kp = [H2S] / [NH4HS]
Given that Kp = 0.110, we can establish the relationship:
Kp = (PH2S) / (PNH4HS)
Since the initial moles of NH4HS is 0.371 moles and the volume of the vessel is 1.00 L, we can calculate the initial partial pressure of NH4HS:
PNH4HS = (0.371 mol) / (1.00 L) = 0.371 atm
Now, we can rearrange the equilibrium constant expression and solve for PH2S:
PH2S = Kp * PNH4HS = (0.110)(0.371 atm) = 0.0408 atm
The equilibrium partial pressure of H₂S is 0.332 atm. The equilibrium partial pressures for PCl₅, PCl₃, and Cl₂ are 0.83 atm, 0.64 atm, and 0.64 atm, respectively.
A.) To determine the equilibrium partial pressure of H₂S for the reaction NH₄HS(s) ⇌ NH₃(g) + H₂S(g) with Kp = 0.110 at 298 K, follow these steps:
Recognize that solid NH₄HS does not affect the equilibrium expression since its activity is 1.Let the equilibrium partial pressures of NH₃ and H₂S be PNH₃ = PH₂S = P since they are produced in a 1:1 ratio.Use the equation for Kp: Kp = PNH₃ × PH₂S = P².Substitute the given Kp value: 0.110 = P².Solve for P: P = √0.110 = 0.332 atm.Therefore, the equilibrium partial pressure of H₂S is 0.332 atm.
B.) For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with Kp = 0.497 at 500 K:
Set up an ICE table. Let the initial pressure of PCl₅ be 1.47 atm, and the changes in pressure be -x for PCl₅ and +x for both PCl₃ and Cl₂.At equilibrium: PCl₅ = 1.47 - x, PCl₃ = x, and Cl₂ = x.The equilibrium expression is: Kp = (PCl₃ × PCl₂) / PPCl₅.Substitute the values: 0.497 = (x × x) / (1.47 - x).Solve the quadratic equation: 0.497 = x² / (1.47 - x).Rearrange to get 0.497(1.47 - x) = x² which gives 0.73 - 0.497x = x².Rearrange into standard quadratic form: x² + 0.497x - 0.73 = 0.Use the quadratic formula x = [-b ± √(b² - 4ac)] / 2a where a = 1, b = 0.497, and c = -0.73.Solve for x to get two possible solutions, but only the positive value is physically meaningful: x ≈ 0.64 atm.Calculate the equilibrium pressures: PPCl₃ = 0.64 atm, PCl₂ = 0.64 atm, and PPCl₅ = 1.47 - 0.64 = 0.83 atm.Therefore, the equilibrium partial pressures are PPCl₅ = 0.83 atm, PPCl₃ = 0.64 atm, and PCl₂ = 0.64 atm.