Unit 1- Atomic Sructure, Amount Of Substances, Bonding, Energetics, Kinetics, Equiliburm And Redox Flashcards

Question

1)

Answer 1

1)no 2)gmol-1 3) 4.2x10^6, 4.20x10^6 4)1000mg 5) 1000000 grams 6) Mr x n

Answer 2

1)mol dm-3 2) Relative atomic mass (Ar), RAM average mass of one atom of an element over one 1/12 mass of one atom carbon 12 3) Relative molecular (formula= ionic formulas) mass Mr or RMM average mass of one molecule over 1/12 mass of one atom of carbon 12 4)molesxAVO 5)atoms/AVO= mass

Answer 3

1)Roman numbers tell the charge on the first atom E.g iron(III) oxide (Fe3+) Fe2O3 -> NEED to balance charges ->use brackets on complex ions only

Answer 4

1)using decimals/fraction in combustion equations 2)doesn’t change state or bonding 3) 1-find the number of moles of the reactant and the mr 2-find the ratio of the reactants 3-multiply the moles the lowest ratio reactant with its moles to find how many moles of the other reactant is needed and compare to the moles first calculated for the limiting reactant (MUST SAY NEED E.G 0.270 MOLES OF HCL) 4)Finding how much of a product is formed: 4-limiting reactant determines how much product is formed 5-multiply moles of limiting reactant by the Mr of the product to get the overall weight in grams

Answer 5

1)6.022x10^23 number of atoms in (1 mole) 12g of carbon 12 ->atoms in a certain mass (molesxAVO) 2)given the atoms then do atoms/ AVO= mass

Answer 6

1)all peaks caused by positive ions 2)mass spec of compounds have M+ parent ion peak (big peak on RHS=mr the mass of the whole molecule) 3)M+1 peak caused by isotopes because carbon and hydrogen isotopes

Answer 7

1)Empirical formula is the simplest whole number ratio of atoms of each element in a compound. 2)Molecular formula is the actual number of atoms of each element in a compound

Answer 8

1)vacuum otherwise air particles would ionise and register on the detector 2) Electrons are lost or gained. Two of these: electron impact and electrospray ionisation

Answer 9

1)elements and substances with low formula mass. why: Electron impact can cause larger organic molecules to fragment 2) 1-Vaporised sample is injected at low pressure 2-An electron gun fires high energy electrons at the sample 3-This Knocks out an outer electron 4-Forming positive ions with different charges E.g. Ti->Ti+ + e–

Answer 10

1) larger organic molecules. The softer conditions of this technique mean fragmentation does not occur 2) • The sample is dissolved in a volatile, polar solvent • injected through a fine hypodermic needle giving a fine mist or aerosol • the tip of needle has high voltage • at the tip of the needle the sample molecule, M, gains a proton, H+, from the solvent forming MH+ • M(g) + H+ -> MH+(g) • The solvent evaporates away while the MH+ ions move towards a negative plate

Answer 11

1) ->Positive ions are accelerated by an electric field ->To a constant kinetic energy ->causing acceleration due electric field so have same KE 2) Given that all the particles have the same kinetic energy, the velocity of each particle depends on its mass. Lighter particles have a faster velocity, and heavier particles have a slower velocity.

Answer 12

1)•The positive ions with smaller m/ values will have the same kinetic energy as those with lighter m/ and will move faster. •The heavier particles take longer to move through the drift area. •The ions are distinguished by different flight times

Answer 13

1)The ions reach the detector accept electrons from it and generate a small current, which is fed to a computer for analysis. 2)The current is produced by electrons transferring from the detector to the positive ions. The size of the current is proportional to the abundance of the species 3)For each isotope the mass spectrometer can measure a m/z (mass/charge ratio) and an abundance

Answer 14

1)distance = speed × time and KE = ½mv2. 2)t = time of flight (s) d = length of flight tube (m) v = velocity of the particle (m s-1) KE = kinetic energy of particle (J) m = mass of the particle (kg) v = velocity of the particle (ms-')

Answer 15

->multiply by 1000 to go from grams to KG

Answer 16

1) given time of one isotope, need to work out time of the other isotope

Answer 17

1) size of the current gives a measure of the number of ions hitting the plate.

Answer 18

1) time of flight is proportional to the square root of the mass of the ions. 2) time taken to travel a fixed distance is then used to find the mass of each ion in the sample. 3)negatively charged metal plate as opposites attract and accepts electrons which generates a current

Answer 19

1) time of flight ->find the abundance and mass of each isotope in an element allowing to determine its relative atomic mass ->find the relative molecular mass of substances made of molecules. 2) small peaks present around the molecular ion peak due to molecular ions that contain different isotopes 3) fragments caused by the break up of molecular ion.

Answer 20

1) simplest whole number ratio of atoms in a compound 2) 1-find the moles of each element (mass/RFM) 2-Find the ratio of whole number (smallest) Moles/smallest moles 3-Round up and present formula E.G: C2H4Br 3)actual number atoms present and type 4) 1-RFM of empirical formula 2-divide the Given RFM by the empirical RFM to get the multiplier E.g x2 C2H4Br -> C4H8Br2

Answer 21

1)you need the RFM to confirm that molecular and empirical formula is the same 2)

Answer 22

1) Maximum theoretical mass-> 1-find moles limiting reactant 2-use ratio of desired product to limiting reactant to find moles of product 3-multiply moles by mr of products to find grams of theoretical mass 2)

Answer 23

1)Work out the empirical formula of magnesium oxide by taking a known mass of magnesium and react it completely with oxygen into a known mass of magnesium oxide. 2)The empirical formula is the formula of a compound that has been worked out by practical experiments E.G: combust a hydrocarbon and find that you get twice as much water vapour as carbon dioxide. ◦ empirical formula is CH4 (remember H2O contains two H atoms). 3) 1-Curl the magnesium 2-Measure crucible +lid and crucible + lid + magnesium 3-heat the magnesium 4-bright light produced and white powder 5-Measure mass of crucible + lid + magnesium oxide after heating

Answer 24

1) Relative atomic mass (Ar) is the average mass of an element compared to one-twelfth the mass of a single carbon-12 atom. 2) 1amu is defined as one-twelfth of the mass of a single carbon-12 atom. Ar= is the average mass of an element in units of amu. 3) The average mass is the average of the masses of the elements, weighted according to their abundance.

Answer 25

1) average mass of a molecule compared to one-twelfth the mass of a single carbon-12 atom. -> 1/12 moles of carbon-12 in 1g of carbon-12 ->Mr is the average mass of a molecule in the units of amu.

Answer 26

1) 6.02 ×1023 Avogadro's constant mole is set so that a mole of carbon-12 atoms weighs 12 grams (12 amu) 2)Number of particles = number of moles × Avogadro's constant 3)mole converts between amu and grams: 1 g = 1 mol × 1 amu 4) Density= mass/volume

Answer 27

1)Dm^3 2)moles ÷ (volume)dm3.= concentration (moldm-3.)

Answer 28

1) Pressure × volume = number of moles × gas constant × temperature (pV = nRT) 2) R is a constant - its value is always 8.314. 3) .Pressure must be in Pascals (Pa). .Volume must be in cubic metres (m3). .Temperature must be in Kelvin (K). 4) Volume= molesx 24

Answer 29

1) Ionic solids will dissolve in solution and will split into their constituent ions. 2) reactions in solution. 3) charge on either side of the ionic equation should be equal. 4)Aq= (substances dissolved in water) 5)The density of water is 1 gcm-3

Answer 30

1)100% 2) 1-It might be difficult to fully separate the product from the reaction mixture. 2-Unexpected reactions (side reactions) can happen. meaning that the intended product is not created. 3-If a reaction is carried out in water, any extracted product may be 'wet', needs to be extracted. 4-incomplete reaction and reversible reaction some of the products have turned back into reactants. 5-The reaction is very slow and hasn't been given enough time to happen. 6-chemicals will remain stuck to the glassware and will be wasted.

Answer 31

1) 1-Conduct the reaction in a conical flask. 2-Add a bung to the top of the flask after adding the reactants. 3-Connect the bung to a syringe so that the gas is collected in the syringe 4-The syringe will allow the volume of gas to be measured 2) • Some hydrogen might have escaped if the bung was not fixed tightly. • Some hydrogen may have escaped between starting the reaction and fixing the bung. • Some hydrogen may have remained in the conical flask and not entered the syringe. • Not all the HCI reacted meaning that not all the possible hydrogen was produced.

Answer 32

1) ->Using the smallest possible amount of reactants saves money. ->higher atom economy means that less waste is produced.

Answer 33

1) Crystallisation is a separation technique used to separate dissolved solids from a liquid. .tripod, gauze, evaporating basin, bunsen burner. 2) 1) Filter the solution to remove any insoluble solids. 2) Heat the solution to boil off about half of the liquid. 3)Leave the remaining solution to cool and evaporate over time. 4)When the crystals have formed, pat them with tissue or paper towel to dry them.

Answer 34

1) .Hydrated copper sulfate crystals. The crystals need some water in it. .solution is boiled dry= anhydrous copper sulfate. .white solid while hydrated copper sulfate is blue.

Answer 35

1)-1 Relative molecular mass because a proton was added

Answer 36

1)-1 Relative molecular mass because a proton was added

Answer 37

1) 1st I.E-> energy required to remove 1 mole of e- from 1 mole of gaseous atoms X(g)->x+(g)+e- 2nd I.E-> X(g)->x2+(g)+e- 2) 3) -> As taking electrons away from positive ions means more energy is required 4) ->new shell closer to nucleus so held more tightly more electrostatic attraction between electrons and nucleus ->LOOK for the big jump

Answer 38

1) ->shielding from shell reduce attraction not held very tightly to the nucleus so very reactive 2) ->easier to take E- as the outer E- is more shielded by other inner shells ->greater distance outer E- nucleus further away so is less attracted to the nucleus

Answer 39

1) Avogadro law-> equal volume different gases at the same pressure and temperature will contain equal number of particles 2) Standard temperature 273 k and pressure 100 kph at standard temperature and pressure 1 mole of any gas occupies volume of 22.7 dm^3. This is a molar volume. 3)PV=nRT 4) P=pressure (PA) R= gas constant (Jk-1 mol-1) V=volume (m^3) n= moles number Temperature= K 5) A gas that obeys this law is called ideal gas.

Answer 40

1) Gas constant= 8.31 2) KPa -> Pa (x1000) dm3 -> m3 (divide 10^3) Cm3 -> m3 (divide 10^6) C -> K (+ 273)

Answer 41

1) is used if the species is ionic 2) mass of one particle x avogardos constant (mass of one mole substance) 3)concentration= mol dm-3 Volume= dm3 4) moles= concentration x volume 5) 1 dm3

Answer 42

1) total mass of all atoms in the product is the same as that of all the atoms in the reactants 2)mass you get compared with the maximum you thought to get 3)measure how much of the products are useful 4)24 dm3 mol-1 or 0.024 m3 mol-1 5)22.4 dm3 mol-1 or 0.0224 m3 mol-1 (273k and 1.013 x 10^5 Pa) 6) 1.013x10^5 Pa 7)1 dm3 or 1x10^-3 m3 8) volumes that are in a simple ratio to each other and to that of any gas products ->all volumes must be measured at the same temperature and pressure

Answer 43

1) gas syringe, victor meyer, Dumas bulb 2)temperature and pressure 3) 4)

Answer 44

1) 1 litre to 1 dm^3

Answer 45

1) ->hard to see a colour change 2)

Answer 46

• burette-> used for sodium hydroxide solution • stand and clamp • 25 cm3 pipette • pipette filler-> used to measure acid • two 250 cm3 conical flasks • two 250 cm3 beakers • funnel • white tile • wash bottle • phenolphthalein indicator • bottle of dilute hydrochloric acid • bottle of sodium hydroxide solution

Answer 47

1. Set up your burette. 2. Use a small volume of the dilute hydrochloric acid to rinse the burette before filling it with the dilute hydrochloric acid a couple of cm3 past 0. run a small amount of the dilute hydrochloric acid through the burette to get the level to exactly 0.00 cm3. 3. Rinse a conical flask with de-ionised water. 4. Rinse a 25 cm3 pipette with the sodium hydroxide solution from the labeled beaker and then, using a pipette filler, pipette exactly 25.0 cm3 of sodium hydroxide solution into a conical. 5. Add a few drops of phenolphthalein indicator to the sodium hydroxide solution in the conical flask and check the colour of the indicator. 6. Record the initial burette reading in the results table make sure that it is to 2 decimal places ending in either a 0 or a 5 7. Titrate the contents of the conical flask by adding dilute hydrochloric acid to it from the burette. Placing a white tile under the conical flask will help you see any colour change clearly. Add the dilute hydrochloric acid bit-by-bit quite fast at first swirling the conical flask the solution. As the colour change becomes more apparent start adding the dilute hydrochloric acid dropwise until the indicator undergoes a definite colour change. 8.wash down the sides of your conical flask with de-ionised water from a wash bottle to ensure you return the hydrochloric acid stuck to the sides to the flask to ensure an accurate end point tire reading. 9. Record the final burette reading in your table of results and calculate the volume of dilute hydrochloric acid used. 10. Repeat the titration until you obtain two results, which are concordant. within 0.10 cm3 of each other.

Answer 48

1) Energy levels 2) Sub shells (sub levels) 3) Where electrons are likely to be found S orbital P orbital 4) electrons enter lowest energy shell 5) Every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied and all electrons in singly occupied orbitals have the same spin. 6)

Answer 49

1) The power number is the number of electrons and the period is the big number 2) opposite directions in the same block but in separate blocks it can be any direction.