atoms Flashcards

1
Q

What is the structure of atoms? and what are there properties?

A
  • Proton’s have a relative charge of +1, have a mass of 1 and are located in the nucleus.
  • Neutron’s have a relative charge of 0, have a mass of 1 and are located in the nucleus.
    -Electron’s have a relative charge of -1, have a mass of 1/1840 and are located in the shells.
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2
Q

How are element’s set out? fact’s about the periodic table?

A

The atomic number of an element is equivalent to the proton’s which is equivalent to the electron’s . the relative atomic mass also know as the moler mass is equivalent to proton - neutrons. and any element’s that are in group 7 come in pair and they are called diatomic.

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3
Q

What are ion’s?

A

Ion’s come from atoms. metal atom’s lose electron’s to become a positive charge and non-metal gain electron’s to become a negative charge.

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4
Q

What are isotope?
Q. What are the similarities and differences(79Br and 81Br)?

A

[Atom’s of the same element that have the same number of proton’s and electron’s but have a different number of neutron’s] They also have the same chemical properties as the same element’s.
A. 79Br and 81Br are similar with the same amount of proton’s and electron’s they have but are different in that 81Br has two more neutron’s than 79Br.

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5
Q

what is relative atomic mass (Ar) / Moler mass?
Q. chlorine gas has a Rm of 35.5

A

[The average mass of an atom relative to the mass of one twelfth of a carbon-12 atom]
A. chlorine atom mass is 35.5 times heavier than 1/12 of carbon-12

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6
Q

What is relative molecular mass (Mr) and how is it calculated?
Q. C6H1206

A

[The mass of a molecule relative to the mass of one twelfth of a carbon-12 atom]
A. C=12X6=72 H=1X12=12 O=16X6=96
72+12+96=180g/mol

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7
Q

What is the relative formula mass? How does it differ from Molecular mass? How is it worked out?
Q. Ca(NO3)2

A

[The mass of a molecule relative to the mass of one twelfth of a carbon-12 atom]. Relative formula mass is the term used when a compound has a giant structure.
A. Ca N2 O6
Ca=1x40.1=40.1 N=2x23=46 O=6x16=96
40.1+46+96=182.1g/mol

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8
Q

What is the relative isotopic mass? what is it need for? and how is it calculated?

A

[The mass of one atom of a particular isotope relative to the mass of one twelfth of carbon-12 atom]. Relative isotopic mass is need for calculating the relative atomic mass also the abundances in also needed (usually %)
mass x percentage of mass/100 + mass x percentage of mass/100
is the abundance is a fraction than it is 1/? x mass of element

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9
Q

What is the mole?

A

[One mole of a substance is the mass that has the same number of particles are there are in exactly 12g of carbon-12]
Mole = 6.02x10^23.

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10
Q

When is Avogadro’s constant (Na) needed? and how is it worked out?

A

When a question have atom’s, ion’s or molecule’s involved you have to use Avogadro’s constant.
Number or Particle’s = No of mole’s X Na (6.02x10^23)

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11
Q

What is moler mass? and how is it worked out?
Q. M(Cl-35)
M(CaC03)

A

[The mass of one more of a substance] units are g/mol-1
A. M(Cl-35) = 35 g/mol-1
M(CaC03) = 100.1 g/mol-1

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12
Q

Q. Solid sulfur exists as a lattice of S8 molecules. Each S8 is a ring of 8 atoms. How many atoms of sulfur are there in 0.0120 mol of s8 molecules.

A

A. 0.0120 x 6.02x10^23=7.224 x 10^21 of S8 7.224 x 10^21 x 8 = 5.7792 x 10^22

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13
Q

What is empirical formula and how is it calculated?
Q. 0.477 g of Oxide iron was reduced. 0.345 g of iron was produced. calculate the empirical formula.

A

[the simplest whole number ratio of the number of atoms of each element in a compound]
A. Fe + O –> FeO 0.345 g + 0.132 g –> 0.477 g
Fe O
0.345/55.8 0.132/16
=0.006/0.006 =8.25x10^-3/0.006
=1 =1.375
1:1.375 x 3 = 3:4 = Fe3O4

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14
Q

What is molecular formula? and how is it worked out?
Q. A hydrocarbon has a empirical formula of CH2 and a relative molecular mass of 112 - what is it molecular formula?
Compound A is a liquid hydrocarbon of relative molecular mass 78 and contains 92.3% carbon. Calculate the molecular formula of A.

A

[Shows the total number of atoms of each element present in a molecule of the compound]
A. 14g/mol x 8 112g/mol (14g/mol is found by adding together the elements Moler mass)
CH2 –> C8H16 112 / 14 = 8
1 x 8 = 8 2 x 8 = 16
100-92.3=7.7%
H C
7.7% 92.3
/ 1 / 12
= 7.7 = 7.7
= 1 = 1 CH (13g/mol)(13 x 6) –> C6H6 (78) (78/13 = 6)

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15
Q

What is water crystallisation? what does anhydrous mean?

A

crystals of some ionic salts contain water loosely bonded to the ions in the crystal lattice. when is crystal is heated it becomes anhydrous which means it does not contain water of crystallisation.

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16
Q

Q. 1.173g of hydrated copper chloride crystals were heated until a constant mass was obtained. 0.641 g of residue remained. Calculate the number of moles of water of crystallisation in the formula: CuCl2xH2O?

A

A. CuCl2xH20 –> CuCl2 + xH2O 1.173 g - 0.641 g = 0.532 g
1.173 g 0.641 g 0.532 g H2 = 1 x 2 = 2 O = 16 2 + 16 = 18
/134.5 /18 Cu = 63.5 Cl = 35.5 x 2 = 71 63.5 + 71 = 134.5
=4.765x10^-3 =0.0295
0.0295/4.765x10^-3 = 6.197
X = 6

17
Q

Q. hydrated salt has the formula of X2CO310H20. On heating the 2.86 g of the hydrated salt, 1.06 g of the anhydrous salt was formed. What is X?

A

A. X2CO310H2O –> X2CO3 + 10H2O 1 : 10
0.01 moles 0.1 moles
2.86 g –> 1.06 g + 1.8 g H = 1 x 2 O = 16 2 + 16 = 18 1.8/18 = 0.1 mole’s
0.01 mole’s of X2CO3
M = m/n M = 1.06/0.01 = 106 g/mol
X2CO3 = 106 g/mol
C = 12 O = 16 x 3 = 48 12 + 48 = 60
106 - 60 = 46
46/2 = 23
X = Sodium.

18
Q

Q. A student heats 12.41 g of hydrated sodium thiosulfate, Na2S2O3.5H2O, to remove the water of crystallisation. a white powder call anhydrous sodium thiosulfate forms.
calculate the expected mass of the anhydrous mass of sodium thiosulfate that forms? (Na2S2O3.5H2O)

A

A. Na2S2O3.5H2O –> Na2S2O3 + 5H2O
12.41 g / 248.2 g
n = 0.05 mole’s 0.05 mole’s 0.25 moles
1 : 1 : 5
Na = 23 x 2 = 46 S = 32.1 x 2 = 64.2 O = 16 x 3 = 48
m = n x M 0.05 x 158.2 = 7.91 g

19
Q

What is the relative charge relative mass and position of proton’s, neutron’s and electron’s?

A

P - +1 - 1 - nucleus
N - 0 - 1 - nucleus
E - -1 - 1/2000 - shells

20
Q

Two isotopes of carbon 12C and 13C
Q. state what is meant by the term isotope’s
isotopes of carbon have the same chemical properties
explain why

A

A. element with the same number of proton’s and a different number of neutron’s
same number of proton’s and electron’s

21
Q

isotope - proton’s - neutron’s - electron’s (118Sn)
in terms of sub-atomic particle’s, how would atom’s of 120Sn differ from atom’s of 118Sn

A

50 - 68 - 50
120Sn has 2 more neutron’s than 118Sn

22
Q

Define the term relative isotopic mass.

A

the mass of one atom of a particular isotope relative to have the mass of one twelfth of carbon-12 atom

23
Q

isotope - relative isotopic mass - abundance (%)
151Eu - 151.0 - 47.77
153Eu - 153.0 - 52.23
calculate the relative atomic mass of the europium sample.
give your answer to two decimal places.

A

151 x 47.77/100 + 153 x 52.23/100
7213.27/100 + 7991.19/100
72.1327 + 79.9119 = 152.0446
152.04 g/mol

24
Q

one coin has a mass of 5.00g and contains 84.0% of copper, by mass calculated the number of copper atoms in one coin
give your answer in standard form and to three significant figures.

A

n = m/M
n = (5 x 84/100) / 63.5 = 0.07 moles
0.07 x (6.02 x 10^23) = 3.98 x 10^23

25
Q

calculate the amount of substance, in mol, in the following
1 . 6.00mg of HF
2 . 0.0150kg of C6H12O6
3 . 3.45 x 10-2 g of Ca(OH)2

A

1 . 6 / 1000 = 0.006 / 20 = 3 x 10-4 n=m/M
2 . 0.0150 x 1000 = 15 C6H1206 = 180
15/180 = 0.083
3. 3.45 x 10-2 / 74.1 = 0.00047 Ca(OH)2 = 74.1

26
Q

calculate the mass in g, of the following
1 . 0.150 mol of HNO3
2 . 0.0500 mol of H3PO4
3. 4.55 x 10-3 mol Ca(NO3)2

A

1 . 0.150 x 63 = 9.45 g HNO3 = 63 m = n x M
2 . 0.0500 x 98 = 4.9 g H3PO4 = 98
3. 4.55 x 10-4 x 164.1 = 0.75 g Ca(NO3)2 = 164.1 g

27
Q

calculate the molar mass on gmol-1 of the following substances:
1 . 5.00 mol of A has a mass of 140 g
2 . 0.125 mol of B has a mass of 9.25 g
3 . 4.50 x 10-2 of C has a mass of 3.825 g

A

1 . 140 / 5.00 = 28 g/mol M=m/n
2 . 9.25 / 0.125 = 74 g/mol
3 . 3.825 / 4.50 x 10-2 = 8.5 x 10-3 g/mol

28
Q

a student decided to investigate the number of carbon atom’s in a ‘pencil lead’. He found that the mass of the ‘pencil lead’ was 0.321 g
calculate the amount, in mol, of carbon atom’s in the student’s pencil lead.
assume that the ‘pencil lead’ is pure graphite.

A

n = m / M n=0.321 / 12 = 0.027
= 0.027 mol

29
Q

a bronze-age shield found on Dartmoor contained 2.08kg of tin.
calculate the number of tin atoms in this bronze shield
give your answer to three significant figures

A

2.08kg = 2080 g / 118.7 = 17.5 mols
17.5 x 6.03 x 10^23 = 1.05x10^25

30
Q

A student reacted 1.44 g of titanium with chlorine to form 5.70 g of a chloride X
1 . how many mole’s of Ti atom’s were reacted
2 . how many mole’s of Cl atom’s were reacted

A

1 . 144/47.9 = 0.03 mole’s n = m/M
2 . 5.70 - 1.44 = 4.26 / 35.5 = 0.12 mole’s