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marshall27 [118]
3 years ago
11

Why is it harder to remove an electron from fluorine than from carbon, or, to put it another way, why are the valence electrons

of fluorine more strongly bound than those of carbon? 1. Carbon has a lower atomic mass than does fluorine. 2. The statement is false; it takes very nearly the same energy to remove an electron from (ionize) both elements. 3. Fluorine has a nearly filled octet, which is always more stable than a partially filled octet. 4. Fluorine has more valence elctrons than does carbon. 5. The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly.
Chemistry
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

5. The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly.

Explanation:

Both fluorine and carbon are located in the second period of the periodic table, it means that they have 2 shells, so the valence electrons are found at about the same distance from their respective nuclei.

But fluorine has a higher atomic number, 9, than the carbon, 6. The atomic number represents how many protons there are in the nucleus, then there are more protons (positive charge) at the fluorine nucleus, and because of that, the attraction force between the nucleus and the valence electron is stronger in fluorine.

If the force is stronger, it will be necessary more energy to break the bond, so it will be harder to remove an electron from fluorine than from carbon.

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How many moles of NH3 can you make from 0.37 moles of H2
uranmaximum [27]

Answer:

The correct answer is 0.25 moles NH₃

Explanation:

First, we need to know the chemical equation for NH₃ formation from H₂. The balanced chemical reaction involved in NH₃ obtention is the following:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

According to this, 2 moles of NH₃ are formed from 3 moles of H₂. We can write that estequiometrical relation as: 2 moles NH₃/ 3 moles H₂.

From the problem, we have to calculate how many moles of NH₃ are produced from 0.37 moles H₂. So, we can simply multiply the number of moles of H₂ to obtain by the convertion factor:

0.37 moles H₂ x 2 moles NH₃/ 3 moles H₂= 0.246 moles NH₃ ≅ 0.25 moles NH₃

Thus, 0.25 moles of NH₃ will be obtained.

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3 years ago
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Butadiene c4h6 reacts with itself at 250 degrees celcius to form a dimer with the formula c8h12. The reaction is second order in
goldfiish [28.3K]

Answer:

<em>Rate</em> = k * [C₄H₆]²

Explanation:

It is possible to write the reaction as:

  • 2C₄H₆ → C₈H₁₂

The differential rate law for a simple second order reaction of the type 2A → B is:

  • <em>Rate</em> = k * [A]²

With the above information in mind, the rate law for the reaction of butadiene would be:

  • <em>Rate</em> = k * [C₄H₆]²
8 0
3 years ago
What is the molarity of a solution prepared by dissolving.
ser-zykov [4K]

Answer:

The molarity (M) of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.

6 0
2 years ago
How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +
Dominik [7]

1) Write the chemical equation.

CH_4+2O_2\rightarrow CO_2+2H_2O

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: <em>unknown</em>.

3) Moles of CH4.

<em>3.1- Set the equation.</em>

PV=nRT

<em>3.2- Plug in the known values and solve for n (moles).</em>

(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=n=0.24\text{ }mol\text{ }CH_4

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: <em>unknown</em>.

<em>5.1- Set the equation.</em>

PV=nRT

<em>5.2- Plug in the known values and solve for V (liters).</em>

(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}V=3.98\text{ }L

3.98 L of O2<em> is required to react with 2.0 L CH4.</em>

.

6 0
2 years ago
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