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A student obtained the following data for the rearrangement of cyclopropane to propene at 500 °C. (CH2)3(g)CH3CH=CH2(g) [(CH2)3], M 0.128 6.40×10-2 3.20×10-2 1.60×10-2 time, min 0 14.4 28.8 43.2 (1) What is the half-life for the reaction starting at t=0 min? min What is the half-life for the reaction starting at t=14.4 min? min Does the half-life increase, decrease or remain constant as the reaction proceeds? ___ (2) Is the reaction zero, first, or second order? _ (3) Based on these data, what is the rate constant for the reaction? min-1

Answers

Answer 1

Answer:

Explanation:

CH2)3(g)CH3CH=CH2(g) [(CH2)3], M time, min

0.128 0

6.40×10-2 14.4

3.20×10-2 28.8

1.60×10-2 43.2

(1) What is the half-life for the reaction starting at t=0 min? min

Half life is the amount of time required for a substance to decay by half of it's initial concentration.

Starting form 0, the initial concentration = 0.128

After 14.4 mins, the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=0min

What is the half-life for the reaction starting at t=14.4 min?

Starting form 14.4min, the initial concentration = 6.40×10-2

After 14.4 mins (28.8 - 14.4), the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=14.4min

Does the half-life increase, decrease or remain constant as the reaction proceeds?

The half life is a constant factor, hence it remains constant as the reaction proceeds.

(2) Is the reaction zero, first, or second order?

Because the half life is independent of the concentration, it is a first order reaction.

In a zero order reaction, the half life Decreases as the reaction progresses; as concentration decreases.

In a first order reaction, the half life Increases with decreasing concentration.

(3) Based on these data, what is the rate constant for the reaction? min-1

The realtionship between the half life and rate onstant is;

k = 0.693 / half life

k = 0.693 / 14.4

k = 0.048125 min-1

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Classify each of the following as a compound, a homogeneous mixture, or a heterogeneous mixture: (a) orange juice; (b) vegetable soup; (c) cement; (d) calcium sulfate; (e) teas

A certain first-order reaction has a rate constant of 2.75 10-2 s−1 at 20.°c. what is the value of k at 45°c if ea = 75.5 kj/mol? webassign will check your answer for the correct number of significant figures. 0.0352 incorrect: your answer is incorrect.

Answers

With an activation energy $($E_a$)$ of 75.5 kJ/mol, the rate constant k for a first-order reaction at 20°C is 2.75 × 10⁻² s⁻¹. At 45°C, k is approximately 0.095 s⁻¹, determined using the Arrhenius equation.

The Arrhenius equation relates the rate constant k, temperature T, activation energy $($E_a$)$ , and the gas constant R:

$k = Ae^{-(E_a)/(RT)}$

Given that $k_1 = 2.75 * 10^(-2) \, \text{s}^(-1)$ at $T_1 = 20^\circ \text{C} = 293.15 \, \text{K}$ and $E_a = 75.5 \, \text{kJ/mol}$ , we want to find $k_2$ at $T_2 = 45^\circ \text{C} = 318.15 \, \text{K}$ .

First, let's find the value of A using the Arrhenius equation at T_1:

$2.75 * 10^(-2) = A e^{-((75.5 * 10^3))/((8.314)(293.15))}$

Solving for A:

$A \approx 3.65 \, \text{s}^(-1)$

Now, use the Arrhenius equation at $T_2$ to find $k_2$ :

$k_2 = (3.65) e^{-((75.5 * 10^3))/((8.314)(318.15))}$

Calculate $k_2$ .

$k_2 \approx 0.095 \, \text{s}^(-1)$

Therefore, the value of k at $45^\circ \text{C}$ is approximately $0.095 \, \text{s}^(-1)$ .

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Final answer:

To find the new rate constant of a first-order reaction under different temperature conditions, we can use the Arrhenius equation, which relates the rate constant, activation energy, and temperature of a reaction.

Explanation:

The student is interested in finding the value of the rate constant (k) at a different temperature for a first-order reaction. The answer can be found using the Arrhenius equation, which defines the relationship between the rate constant (k) of a reaction and the temperature at which the reaction occurs. The activation energy (Ea) is also necessary.

The Arrhenius equation is: k = A * exp(-Ea/(R*T)), where A is the pre-exponential factor, R is the universal gas constant (the value of R should be 8.314 J/mol.K to match the Ea units), and T is the temperature in Kelvin.

At the first condition, you have the value of k and the corresponding T (convert Celsius to Kelvin by adding 273.15). With these values and the known Ea, you can solve for A. Then, using the value of A, Ea, and the second T (also converted to Kelvin), you can solve for the new k.

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Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.

Answers

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

$\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} * 100 \%$

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

$\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} * 100 \%\n\n\text{Percent Mass} = 10.6 \%}$

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

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Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes 52.94 g = 63.55 g

1 Sulphur 32.07 g =

32.07 g 2 Hydrogen is = 2.02 g

4 Oygen =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol 18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

Which one of the following combinations cannot function as a buffer solution?A) HCN and KCN
B) NH3 and (NH4)2SO4
C) HNO3 and NaNO3
D) HF and NAF
E) HNO2 and NaNO2

Answers

Answer:

it is B

; )

Which one of the following combinations cannot function as a buffer solution?

A) HCN and KCN

B) NH3 and (NH4)2SO4

C) HNO3 and NaNO3

D) HF and NAF

E) HNO2 and NaNO2

How many water molecules are in a block of ice containing 1.50 mol of water (H2O)?

Answers

1.50 moles H20(6.02*10^23 molecules/1 mole)
9.03*10^23 molecules

Final answer:

The number of water molecules in a 1.50 mol block of ice is calculated by multiplying the number of moles of water by Avogadro's number. The result is approximately 9.033 x 10^23 water molecules.

Explanation:

In chemistry, the amount of substance in moles is related to the number of particles (atoms, molecules) through Avogadro's number. Avogadro's number, which is 6.022 x 1023 particles/mol, tells us the number of molecules in one mole of a substance.

To calculate the number of water molecules in 1.50 mol of water, you would multiply the number of moles of water by Avogadro's number:

1.50 mol of water x 6.022 x 1023 water molecules/mol of water = 9.033 x 1023 water molecules

Therefore, there are approximately 9.033 x 1023 water molecules in a 1.50 mol block of ice.

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Secondary amines react with the nitrosonium ion to generate ________. n-nitrosoamines diazonium salts anilines imines oximes

Answers

Answer:
Secondary amines react with the nitrosonium ion to generate N-Nitrosoamines.

Explanation:
Nitosonium Ion is generally utilized in the formation of Diazonium Salts which are considered excellent starting Material from synthesis point of View. Diazonium salts are formed by reacting Primary Amine or Anilines with Nitrosonium Ions. In our case, the Amine given is Secondary. So, reaction of Sec. Amines with Nitrosonium Ions stops after the formation of N-Nitrosoamine as there is no Hydrogen attached to Nitrogen atom of Amine to be eliminated and form a double and eventually triple bond with the Nitrogen atom of Nitrosonium Ion.

Answer the following questions about the last lesson below. What did we figure out about chain reactions? What did we figure out about fission? Fission produce more energy than a typical fuel sources, chain reaction

Answers

Answer:

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