All categories

3 years ago

Why are fluorine (F), chlorine (Cl), and iodine (I) in the same group of the periodic table?

Chemistry

2 answers:

JulsSmile [24]3 years ago

5 0

Answer:

Explanation:

Verdich [7]3 years ago

3 0

Fluorine (F), Chlorine (Cl) and Iodine (I) are all found in the same group on the Periodic Table because they have similar physical properties. Since they are all Halogens, they have 7 valence electrons in their outer shell. In order to get a total of 8, they naturally combine with elements of the same isotope (itself), so D comes close to being correct, but it's not the best answer choice.

You might be interested in

A house is built with a granite countertop. The heat capacity of the countertop is 158.5 kJ/°C. A hot pan of water is placed on

Svetlanka [38]

The temperature of the countertop changes by 0.032 °C

The quantity of heat transferred to the countertop is given by

Q = CΔT where Q = quantity of heat transferred to the countertop = 5000 J = 5 kJ, C = heat capacity of the countertop = 158.5 kJ/°C and ΔT = temperature change of the countertop.

Since we require the temperature change of the countertop, we make ΔT subject of the formula.

So, ΔT = Q/C

So, substituting the values of the variables into the equation, we have

ΔT = Q/C

ΔT = 5 kJ/158.5 kJ/°C

ΔT = 0.032 °C

So, the temperature of the countertop changes by 0.032 °C

Learn more about temperature change here:

brainly.com/question/16384350

8 0

3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$