The correct name for P5O2 is

why is it harder to remove an electron from fluorine than from carbon? to put it another way, why are the outermost electrons of

Verizon [17]

It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.

The energy required to remove an electron from an atom is called ionization energy.

The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.

Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.

Learn more: brainly.com/question/16243729

6 0

2 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

2 years ago

Calculate the mass of a sample of lead (cPb = 0.16 J/g℃) when it loses 200 J cooling from 75.0℃ to 42.0℃.

egoroff_w [7]

Looses 299 please collins 75.0

7 0

2 years ago

If 1.546 g of copper was used by a student at the start of the lab, and 0.732 g of copper were obtained at

mash [69]

Answer: Percent recovery is 47.34 %

Explanation:

Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

${\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100$

Putting in the values we get:

${\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%$

Therefore, the percent recovery is 47.34 %

8 0

2 years ago

How could your school reduce the amounts of municipal soild waste it produces ? Inculding where you think the most waste is prod

katrin [286]

Reducing solid waste is reducing the amount of garbage that goes into our landfills. These are items we use each day, and then get rid of by putting them into the trash. Solid waste comes from homes, businesses and industries. If you want to reduce solid waste, you need to look at some of the following ways to reduce, reuse and recycle.

6 0

2 years ago