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ra1l [238]
3 years ago
15

What are similarities and differences between ionization energy and electron affinity

Chemistry
1 answer:
Vladimir [108]3 years ago
8 0
Ionization energy: the energy required to remove an electron from a neutral atom. Electron affinity: the energy change when a neutral atom attracts an electron to become a negative ion.
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Explain two ways that bills penguin toy is similar to the water cycle
Alja [10]

All the water on the earth is the same water that has been here since

<h2><u><em>BRAINLIST </em></u></h2>
3 0
3 years ago
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

<h3>pH = 11.52</h3>
5 0
2 years ago
using the equaation 2h2+o2--&gt;2h2o if 10.0g of hydrogen are used in the presence of excess oxygen how many grams of water will
astra-53 [7]

Answer:

90g of H2O

Explanation:

2H2 + O2 —> 2H2O

First, we calculate the molar masses of H2 And H20.

Molar Mass of H2 = 2g/mol

Mass conc of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of H2O = 2 + 16 = 18g/mol

Mass conc of H2O from the balanced equation = 2x18 = 36g

From the equation,

4g of H2 produced 36g of H2O

Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O

7 0
2 years ago
A(n) ____________________ is a group of two or more atoms that are held together by chemical bonds. *
Kryger [21]

Answer:

A molecule is a group of two or more atoms that are held together by chemical bonds.

Explanation:

6 0
2 years ago
Determine the oxidation number of each element in these compounds or ions. (a) au2(seo4)3 (gold(iii) selenate) au = se = o = (b)
Stels [109]
<span>                                                    Au</span>₂(SeO₄)₃

                                         O = -2 × 4 = -8
                                             Se  =  + 6
So,
                                            (+6 - 8) = -2

Means (SeO₄) contains -2 charge, Now multiply -2 by 3
                                             
                                             -2 ₓ 3 = -6
Means,
                             Au₂ + (-6) = 0
               
                            Au₂  = +6
Or,
                            Au  =  6 / 2

                            Au  = +3
Result:
                            Au  =  +3
                            Se  =  +6
                            O   =  -2

                                                      Ni(CN)₂


Cyanide (CN⁻) contains -1 charge,
So,
                              N  =  -3
                              C  =  +2
Then,
                                         Ni + (-1)₂  =  0

                                               Ni - 2  =  0
Or,
                                                     Ni =  +2
Result:
                            N  =  -3
                            C  =  +2
                           Ni  =  +2




6 0
3 years ago
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