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Montano1993 [528]
2 years ago
8

Why are the trends and exceptions to the trends in ionization energy observed? Check all that apply. Ionization energy tends to

increase down a group because the electrons get farther away from the nucleus. Ionization energy tends to increase across a period because the nuclear charge increases. Ionization energy tends to increase across a period because electrons are added to the same main energy level. The ionization energies of the elements in Group 16 tend to be slightly smaller than the elements in Group 15 because the fourth electron is added to an unfilled p orbital. The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.
Chemistry
2 answers:
NNADVOKAT [17]2 years ago
5 0
The trends and exceptions to the trends in ionization energy observed includes;
B, ionization energy tends to increase across a period because the nuclear charge increases.
C, ionization energy tends to increase across a period because electrons are added to the same main energy level.
E, The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron, in the p orbital from the nucleus. 
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example to remove an electron from a neutral fluorine atom to form a positively charged ion. <span />
Georgia [21]2 years ago
4 0

answers are

2) Ionization energy tends to increase across a period because the nuclear charge increases.

3)Ionization energy tends to increase across a period because electrons are added to the same main energy level.

5) The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.


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Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli
Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

<em></em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

<em></em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
3 0
2 years ago
Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0
2 years ago
Read 2 more answers
How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a
Sindrei [870]

Answer:

Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.

Explanation:

Hope this helped :)

5 0
2 years ago
The compound AX2 decomposes according to the equation, 2 AX2(g) =&gt; 2 AX(g) + X2(g). In one experiment, AX2 was measured at va
Karolina [17]

Answer:

0.0011 mol/L.s

Explanation:

The average rate of disappearing of the reagent is the variation of the concentration of it divided by the time that this variation is being measured. The reaction rate, is proportional to the coefficient of the substance, so, for a generic reaction:

aA + bB --> cC + dD

rate = -(1/a)Δ[A]/Δt = -(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δdt = (1/d)Δ[D]/dt

The minus sign is because of the reagent is desapering, so:

rate = -(1/2)*(0.0209 - 0.0300)/(10 - 6)

rate = 0.0011 mol/L.s

8 0
2 years ago
(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pus
balandron [24]

Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

5 0
2 years ago
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