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zalisa [80]
3 years ago
6

An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy. True False

Chemistry
1 answer:
andrew11 [14]3 years ago
8 0

The statement "An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy" is false.

In this context , we will define ionization energy as the minimum energy required to remove a valence electron from a neutral atom in it's gaseous state.  In a sense the ionization energy is a measure the amount of 'difficulty' of making something an ion. A high ionization energy means that it takes a lot of energy to remove a valence electron from that atom. A low ionization energy means that it is easy to remove a valence electron from the atom. It is known that group 1 elements generally have a low ionization energy.  On the other hand, it is harder for noble gasses and group 7 atoms to loose electrons because they have higher ionization energy.

To form a positive ion, you have to remove an electron. When an electron is removed from an atom, there ion formed has more positive charges than negative charges in it, making it net positive. We have established that atoms with low ionization energy loose elections much more easily. We have also established that atoms with high ionization energy do not loose electrons easily. From this we can gather that the statement is false. An atom with high ionization energy will not form a positive ion more easily that an atom with low ionization energy.

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A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0
3 years ago
Magnesium hydroxide is added to a solution of hydrochloric acid. A reaction occurs and magnesium chloride and water are formed.
kodGreya [7K]

Answer:Magnesium (Mg) is a - reactant

Hydrogen (H2) is a - product

magnesium chloride (MgCI2) is a - product

hydrochloric acid (HCI) is a - reactant

Explanation: It’s in my notes

8 0
3 years ago
What is the pH of a solution that has a [OH-] of 5.08x10^-5 M
sattari [20]

Answer:

The pH of a solution that has a [OH-] of 5.08x10^-5 M is 5

Explanation:

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Which of the following statements about the greenhouse effect is false? Greenhouse gases include carbon dioxide, chlorofluorocar
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