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3 years ago

Which factor determine the reactivity of alkyl halide

Chemistry

1 answer:

Pavel [41]3 years ago

6 0

Answer:

This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.

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Which element could provide one atom to make an ionic bond with fluorine? (Hydrogen will not make ionic bonds easily, so do not

Ainat [17]

Any element from group 1 of the Periodic Table.

6 0

4 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

1. Table

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

2. Orders

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

3. Rate equation

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

4. Rate constant, k

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

5. Rate law:

$rate=k[A]^2C=0.0075[A]^2[C]$

6. Substitute

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

7 0

3 years ago

Question 15 (1 point)

Yuliya22 [10]

Answer:

mh cjwe cejwh cjefblewkbfhfwouf

Explanation:

jhbckwjebfljbewfjbwefbwc

6 0

3 years ago

A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.

Semmy [17]

Answer

2.7956 * 10^19 photons

Givens

Wavelength = λ = 525 * 10^-9 meters [1 nmeter = 1*10^-9 meters]
c = 3 * 10^8 meters
E = ???
W = 100 watts
t = 1 second
h= plank's Constant = 6.26 * 10^-34 J*s

Formula

E = h * c / λ

W = E / t

Solution

E = 6.26 * 10^-34 j*s * 3 * 10^8 m/s /525 * 10^-9 (m)

The meters cancel out. So do the seconds. You are left with Joules as you should be.

E = 3.577 * 10^-18 Joules

What you have found is the energy of 1 photon.

Now you have to find the Joules from the watts.

W = E/t

100 * 1 second = 100 joules

1 photon contains 3.577 * 10 ^ - 18 Joules

x photon = 100 joules

1/x = 3.577 * 10^-18 / 100 Cross multiply

100 = 3.577 * 10 ^ - 18 * x Divide both sides by 3.577 * 10 ^ - 18

100/3.577 * 10 ^ - 18 = 3.577 * 10 ^ - 18x / 3.577 * 10 ^ - 18

2.7956 * 10^19 photons = x

7 0

3 years ago

Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/

Ivenika [448]

Answer:

The answer to your question is: C₃H₃O This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

44 g -------------------- 12 g

3.2007 g ------------ x

x = (3.2007 x 12) / 44

x = 0.8729 g of Carbon

12 g of C -------------- 1 mol

0.8729 g -------------- x

x = (0.8729 x 1) / 12

x = 0.0727 mol of Carbon

For Hydrogen

18 g ---------------------- 1 g

1.3102 g ------------------- x

x = (1.3102 x 1) / 18

x = 0.0727 g of Hydrogen

1 g ------------------------ 1 mol

0.0727g ---------------- x

x = (0.0727 x 1)/1

x = 0.0727 mol of Hydrogen

For oxygen

g of Oxygen = g of sample - g of Carbon - g of hydrogen

g of Oxygen = 1.3109 - 0.8709 - 0.0727

g of Oxygen = 0.3673

16 g of Oxygen ------------- 1 mol of O

0.3673 g --------------------- x

x = (0.3673 x 1)/ 16

x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon 0.0727 / 0.023 = 3.1 ≈ 3

Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3

Oxygen 0.0230 / 0.023 = 1

C₃H₃O

8 0

3 years ago

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