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Chemistry > UWORLD CHEM 5/20/2020 > Flashcards

UWORLD CHEM 5/20/2020 Flashcards

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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
Q

Henderson Hasselback

A

pH= Pka+ log conjugate base/ acid

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26
Q

Buffers

A

Weak acid/ conjugate base

Weak base/ Conjugate acid

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27
Q

Buffers meaning

A

They resist pH change

28
Q

Buffers at the equivalent point

A

pka= pH

29
Q

Indicators

A

Weak acids
in the range of that acid
pKa close to it is the indicator

30
Q

High intensity means

A

high concentration and lower pH

31
Q

Signal intensity spectrometry

A

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
Q

Paper chromatography

A

Solvent mobile phase

stationary phase paper

33
Q

How does paper chromatography work?

A

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
Q

Rf

A

the distance traveled by the pigment/ distance of the solvent front

35
Q

Rf=0

A

The object did not move

36
Q

Rf=1

A

The substance moved as far as the solvent from

37
Q

Thin Layer chromatography

A

Advanced paper chromatography

instead of a paper, it has the specially coated substance

38
Q

In HPLC if they are polar

A

The higher the retention time and will go slowers

39
Q

In HPLC

A

polar molecules will move more slowly

40
Q

Recrystallization

A

Barely let your molecule to be dissolved, then let it to crystalize again= compounds need to be pure

41
Q

Recrystallized

A

Cool temp, solubility decreases,

42
Q

Aldose

A

Sugar that contains aldehydes, CHO

43
Q

Pyranose

A

the 6 membrane ring in a cycle

44
Q

Sugars

A

are linear carbohydrates that contain carbonyl and two other bounded to hydroxyl atom

45
Q

Hemiacetyle

A

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
Q

Furanose

A

Sugars that are 5 carbon

47
Q

Hemiketal vs hemiacetal

Furanose vs. Pyranose

A

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
Q

Aldol condensation

A

Because of the acidic alpha proton

49
Q

2 acetaldehyde

A

Aldo

50
Q

oxidations

A

Aldehyde oxidizes to carboxylic acid

Ketones do not oxidize anymore

51
Q

1,3-dicarbonyls or active methylene compounds.

A

1,3-dicarbonyls have 2 carbonyl groups flanking a carbon atom with an acidic proton.

52
Q

Internal H bonding

A

1,3-dicarbonyls: internal H-bonding

53
Q

Tautomerism causes

A

one of the carbonyl changes to one of its enol form, which contains H bonding in the OH bond with the other carbonyl.

54
Q

Enol form

A

is the one with the alcohol

55
Q

Keto form

A

is the one from the ketone

56
Q

Keto form vs enol form

A

keto is more stable

57
Q

Organometallic Compounds

A

Makes R which attacks c=o and makes R-C-OH

58
Q

What is the purpose organometallic compound

A

To make a carbon carbon bond

59
Q

Wolff Kishner

A

Reduces C=O to CH2

60
Q

Sugars can interconvert between linear and cyclic forms

A

In the linear form, the carbonyl carbon (anomeric carbon) undergoes the nucleophilic attack by one of the hydroxyl groups on the sugar cyclize.

61
Q

fRUCTOSE

A

Ketose with the anomeric carbon at C-2

62
Q

epimers

A

glucose and mannose are epimers b/c they have the substituent at C-2

63
Q

epimers

A

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
Q

In sugars R configuration

A

D

65
Q

In sugars S configuration is

A

L

66
Q

alpha anomers

A

opposite sides

67
Q

Beta anomers

A

Same sides

Chemistry (4 decks)

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