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3 years ago

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An alternative definition of electronegativity is Electronegativity= constant (I.E — E.A.) where I.E. is the ionization energy a

nd E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 12 to calculate the (I.E. - E.A.) term for F, Cl, Br, and I. Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kJ/mol, respectively.

1 answer:

Kazeer [188]3 years ago

8 0

Answer:

The electronengativity values of given elements is as follows.

Fluorine - 4

Chlorine -3

Bromine - 2.9

Iodine- 2.5

Explanation:

Electronegativity =consant (I.E-E.A)

The electron affinity and ionization energy values of the given elements is as follows.

(In attachment)

First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".

<u>Fluorine:</u>

$4=constant[1678-(-327.8)]$

$Constant=0.0019942168$

By using this constant values we can find electronegatvity values of remaining elements.

<u>Chlorine:</u>

$Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3$

Therefore, electronegativity of chlorine is 3.

<u>Bromine:</u>

$Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9$

Therefore, electronegativity of bromine is 2.9.

<u>Iodine:</u>

$Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5$

Therefore, electronegativity of iodine is 2.5.

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Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of $O_{2}$ gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of $O_{2}$ (molar mass = 32.0 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol$

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L$

Thus, we can conclude that the volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0

3 years ago

12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?

Zielflug [23.3K]

Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?$

Putting values in above equation, we get:

$0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L$

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3 years ago

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Goshia [24]

Answer:

c

Explanation:

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3 years ago

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas

luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams

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3 years ago

If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?

statuscvo [17]

The process in which the concentration of the solution is lessened by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

$C_{1}V_{1}=C_{2}V_{2}$ (1)

where,

$C_{1}$ is the initial concentration

$V_{1}$ is the initial volume

$C_{2}$ is the final concentration

$V_{2}$ is the final volume

Now,

$C_{1}$ = 0.850 M

$V_{1}$ = 4.12 L

$C_{2}$ =?

$V_{2}$ = 10.00 L

Substitute the give values in formula (1),

$0.850 M\times 4.12L=C_{2}\times 10.00 L$

$C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}$

= $0.3502 M$

Thus, the final concentration of the $H_{3}PO_{4}$ solution = $0.3502 M$

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2 years ago