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Chemistry > UWORLD CHEM 5/20/2020 > Flashcards

UWORLD CHEM 5/20/2020 Flashcards

1
Q

Positron decay

A

Atomic number decreases

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2
Q

Isotopes

A

different numbers of neutrons

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3
Q

Chemical behavior

A

determined by the electron configuration

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4
Q

Physical and chemical properties of the isotope

A

They have a same chemical properties and different physical properties

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5
Q

Suppose that the (p-n) nuclear reaction described in the passage was attempted but resulted in the 18O nucleus absorbing a proton without ejecting a neutron. The resulting nucleus formed would be an isotope of

A

F is the isotope of O18

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6
Q

Radioactive decay

A

the time is required for the amount of isotope to become by half

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7
Q

Fraction remaining

A

1/2^n

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8
Q

Measured Activity

A

1/2^n *( Initial Activity)

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9
Q

molar mass

A

amount of mass compound in 1 mole of the compound

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10
Q

How products nad reactants are related to eachother

A

By the mole ratio

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11
Q

millimoles

A

Multiply by 4

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12
Q

Number of moles

A

Dividing the mass of sample by its molar mass

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13
Q

Electrostatic attraction

A

The van der Waals interaction between partially positive and negative molecules

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14
Q

Core Electrons

A

located between the valence electrons

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15
Q

Lewis Structure

A

Chemical compounds that repented by the dots and shows their valence electrons.

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16
Q

Octet Rule

A

S and P rech the full valence

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17
Q

Exceptions to octet rule

A

Small atoms such as H and He

Larger atoms in Period 3 have vacant d orbitals are utilized

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18
Q

SO3

A

is a bent.

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19
Q

Titrants

A

To find the mole of unknown

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20
Q

cv= mole

A

c= mole/v

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21
Q

Equivalent point

A

Moles of base= Moles of acid

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22
Q

Titrations

A 
1. Strong/ Strong
Strong acid & Strong base
2. Strong /Weak
Strong acid/ weak base pH<7
Strong base/ weak acid pH>7
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23
Q

Polyprotic

A

A substance that can donate more than 2 protons

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24
Q

How many NaOH will take to kill H2A?
500 ml of 0.25 M of H2A
AND 0.25 OF NaOH

A

The first half point according to c1v1=c2v2 is 500.

The second equivalent point is 1000

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25
Henderson Hasselback
pH= Pka+ log conjugate base/ acid
26
Buffers
Weak acid/ conjugate base | Weak base/ Conjugate acid
27
Buffers meaning
They resist pH change
28
Buffers at the equivalent point
pka= pH
29
Indicators
Weak acids in the range of that acid pKa close to it is the indicator
30
High intensity means
high concentration and lower pH
31
Signal intensity spectrometry
The signal intensity of a mass spectrometry peak corresponds to the relative quantity of ions at a given mass-to-charge ratio m/z.  Increased charge yields a decreased m/z ratio; therefore, an increase in the number of multiply charged particles increases signal intensity at lower m/z ratios.
32
Paper chromatography
Solvent mobile phase | stationary phase paper
33
How does paper chromatography work?
Those with the greater affinity with paper will stick to it and those with the lower affinity to paper will wash off and this is about all the dyes
34
Rf
the distance traveled by the pigment/ distance of the solvent front
35
Rf=0
The object did not move
36
Rf=1
The substance moved as far as the solvent from
37
Thin Layer chromatography
Advanced paper chromatography | instead of a paper, it has the specially coated substance
38
In HPLC if they are polar
The higher the retention time and will go slowers
39
In HPLC
polar molecules will move more slowly
40
Recrystallization
Barely let your molecule to be dissolved, then let it to crystalize again= compounds need to be pure
41
Recrystallized
Cool temp, solubility decreases,
42
Aldose
Sugar that contains aldehydes, CHO
43
Pyranose
the 6 membrane ring in a cycle
44
Sugars
are linear carbohydrates that contain carbonyl and two other bounded to hydroxyl atom
45
Hemiacetyle
a hemiketal (ketose) form.  A hemiacetal originates from an aldehyde and consists of a carbon bonded to a hydroxyl group (–OH), an alkyl group, an –OR group (where R is any carbon chain), and a hydrogen atom.
46
Furanose
Sugars that are 5 carbon
47
Hemiketal vs hemiacetal | Furanose vs. Pyranose
a hemiketal (ketose) form.  A hemiacetal originates from an aldehyde and consists of a carbon bonded to a hydroxyl group (–OH), an alkyl group, an –OR group (where R is any carbon chain), and a hydrogen atom.
48
Aldol condensation
Because of the acidic alpha proton
49
2 acetaldehyde
Aldo
50
oxidations
Aldehyde oxidizes to carboxylic acid | Ketones do not oxidize anymore
51
1,3-dicarbonyls or active methylene compounds.
1,3-dicarbonyls have 2 carbonyl groups flanking a carbon atom with an acidic proton.
52
Internal H bonding
1,3-dicarbonyls: internal H-bonding
53
Tautomerism causes
one of the carbonyl changes to one of its enol form, which contains H bonding in the OH bond with the other carbonyl.
54
Enol form
is the one with the alcohol
55
Keto form
is the one from the ketone
56
Keto form vs enol form
keto is more stable
57
Organometallic Compounds
Makes R which attacks c=o and makes R-C-OH
58
What is the purpose organometallic compound
To make a carbon carbon bond
59
Wolff Kishner
Reduces C=O to CH2
60
Sugars can interconvert between linear and cyclic forms
In the linear form, the carbonyl carbon (anomeric carbon) undergoes the nucleophilic attack by one of the hydroxyl groups on the sugar cyclize.
61
fRUCTOSE
Ketose with the anomeric carbon at C-2
62
epimers
glucose and mannose are epimers b/c they have the substituent at C-2
63
epimers
glucose and mannose are epimers b/c they have the substituent at C-2 Isomers with diffrent configurations of atoms around one of several asymmetric carbon atoms present.
64
In sugars R configuration
D
65
In sugars S configuration is
L
66
alpha anomers
opposite sides
67
Beta anomers
Same sides

Chemistry (4 decks)

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