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3 years ago

Chlorine radicals perform the first propagation step:

Chemistry

1 answer:

Black_prince [1.1K]3 years ago

6 0

Answer:

b. radicals form easily in the presence of chlorine radicals.

Explanation:

Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."

This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.

Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."

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Select the correct answer. Which statement is true according to the quantum model of the atom? A. When an electron drops down fr

nalin [4]

D: After they emit light, electrons are sucked into the nucleus of the atom.

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3 years ago

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at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were

Nataly [62]

Answer: The percent change in volume will be 25 %

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1$

Putting values in above equation, we get:

$\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L$

Percent change of volume = $\frac{\text{Change in volume}}{\text{Initial volume}}\times 100$

Percent change of volume = $\frac{(2-1.5)}{2}\times 100=25\%$

Hence, the percent change in volume will be 25 %

5 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$