test Flashcards
when I calc percentage uncertainty
x 2 if ur calc two readings if its just one just leave(usually burette)/divide by diff in mass/percentage uncertainty/the ones given
forming the roman numeral thing do it for fe203 and cl02-
o3=-6 so fe =+6 one fe =+3
so its fe(|||)oxide
o2=-4 to get to -1 cl=+3
so cl3=chlorate and its chlorate(|||)oxide
for half equations what side does the electrons go if its reduction and what is reduction
left side
gain of electrons, decrease in oxidation no and oxidising agent
calcium n water
bariumm n water
Effervescence/fizzing/bubbles
OR
Ca/solid disappears/dissolves OR
Forms a white ppt/solid✓
more vigorous Effervescence/fizzing/bubbles
ba/solid disappears/dissolves faster Forms a white ppt/solid slower
is it a redox of Sr+2H2O→Sr(OH)2 +H2
st-0–+2
H-+2–0
if ur doing oxidayo of HCL u u just use normal number of cl but if its mno2 u calc it
how to reduce uncertainties in a titration
inc titre volume.inc vol n conc of substance I flask.or dec conc of substance in burette
if ur doing
2Sr(NO3)2
do NO3- first
ur doing it to get charge of -1
-2 x3=-6 and get it to -1 so +5
in hess cycle don’t when your trying to make x u don’t have to like swap signs to make x on the same side and usually if ur trying ti fund formation of one thing like 4NO3..
make it like 4x
when we doing a calc question for enthalpy change and we need to find the moles, do it of the limiting reactant eg 14(a).
Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
In the experiment, 3.18 g of Na2CO3 are added to 50.0 g of 2.00 mol dm−3 HCl, an excess. The temperature of the reaction mixture increases by 5.5 °C.
Calculate ΔHr, in kJ mol−1.
Give your answer to three significant figures.
The specific heat capacity, c, of the reaction mixture is 4.18 J g−1 K−1.
Energy
q calculated correctly = 1149.5(J)✔ OR 1.1495 (kJ) ✔
Amount, n, of Na2CO3 calculated correctly= 0.03(00) ✔
−40.8 kJ mol–1 if 53.18 used in calculation of q
ALLOW −40.7 kJ mol–1 if q is rounded to 1220 from 53.18 earlier
what’s a catalyst
Catalyst lowers activation energy
Reaction proceeds via a different route/pathway
when ur trine do oxidation number the one u trine figure out you do the one next to it first
2Cl 2 + 2Ca(OH)2 CaCl 2 + Ca(OCl)2 + 2H2O
show that disproportionation has taken place.
m 0 in Cl2 to +1 in Ca(OCl)2 OR ClO-
Reduction
from 0 in Cl2 to –1 in CaCl2 OR Cl-
less yield produced
more than one termination step
Formation of 1-bromobutane
OR (Br) subsitution in a different position
if given`
✅ The mass or moles of product actually made
✅ The percentage yield
❓ You’re asked to find the mass or moles of reactant needed
(c)
13
2-Bromobutane can also be prepared by reacting butan-2-ol, CH3CH2CHOHCH3, with sodium bromide and sulfuric acid (Reaction 5.3).
CH3CH2CHOHCH3 + H+ + Br– CH3CH2CHBrCH3 + H2O Reaction 5.3 2-Bromobutane is a liquid with a boiling point of 91 °C and does not mix with water. (i) A student plans to prepare 10.0 g of 2-bromobutane using Reaction 5.3.
The percentage yield is 67.0%.
Calculate the mass of CH3CH2CHOHCH3 needed for this preparation. Give your answer to 3 significant figures.
Theoreticalamount=
Percentageyield(asadecimal)
Actualamount
n(2-bromobutane)
= 10.0 = 0.073(0)…. (mol)
136.9 n(CH3CH2CHOHCH3)
= 0.0730…. × 100 = 0.109 (mol) 67.0
mass CH3CH2CHOHCH3
= 0.109 × 74.0 = 8.07 g
barium meal
patient swallows water shaken with barium sulfate, white ppt coats inner
lining of gut, X-ray taken→ can identify abnormalities e.g. ulcers/tumours
determination of enthalpy change epeirmetally combustion
Measure certain volume of water, pour into beaker, record initial temperature
○ Add fuel to spirit burner + weigh
○ Place spirit burner under beaker + light & stir water with thermometer
○ Extinguish flame after about 3 mins + immediately record temp water
○ Re-weigh spirit burner
Experimentally finding formula hydrated salt:
○ Weigh empty crucible, then add hydrated salt & reweigh
○ Use pipe clay triangle to support crucible on a tripod. Heat for 1 min strong, 3
mins gentle
○ Leave to cool then weigh
● Accuracy of experimental formula: assumes all water has been lost (solution= heat to constant mass) & assumes no further decomposition (difficult if no colour change)
how to do purify,,,
Separating funnel to remove organic layer
from aqueous layer
Anhydrous salt to dry organic layer
Distillation to purify the product
how to do max theoretical yield
do normal mass as u would mass/rfm
if its asymmetrical dont forget to look at carbocations when you do elec
combustion reactions always be exo thermic so in those big questios
the end answer always negative
A solution contains 5.14 g of Sr(OH)₂ in 200 cm³ of solution.
Calculate the concentration of OH⁻ ions in mol dm⁻³.
(Give your answer to 3 significant figures.)
Mr of Sr(OH)₂ = 87.62 + 2(16.00 + 1.01) = 121.64 g/mol
Moles of Sr(OH)₂ = 5.14 ÷ 121.64 = 0.0422 mol
[Sr(OH)₂] = 0.0422 ÷ 0.200 = 0.211 mol dm⁻³
Each Sr(OH)₂ → 2 OH⁻ ions
→ [OH⁻] = 2 × 0.211 = 0.422 mol dm⁻³
When hydrated strontium chloride is heated, the water of crystallisation is removed, leaving a
residue of anhydrous strontium chloride.
A student carries out an experiment to find the value of x in the formula of hydrated strontium
chloride, SrCl2xH2O.
The student’s method is outlined below.
Step 1
Weigh an empty crucible.
Add SrCl2xH2O to the crucible and reweigh.
Step 2
Heat the crucible and contents for 10 minutes. Allow to cool and reweigh.
Step 3
Heat the crucible and residue for another 5 minutes. Allow to cool and weigh the crucible and residue.
Repeat step 3 a further two times.
The student’s results are shown below:
Mass of empty crucible/g-15.96
Mass of crucible + SrCl2xH2O/g-18.65
First mass of crucible + residue/g-17.66
Second mass of crucible + residue/g-17.61
Third mass of crucible + residue/g-17.58
Fourth mass of crucible + residue/g-17.58
(i) Calculate the value of x in SrCl2xH2O. Give your answer to 2 significant figures.
n(SrCl2) = 1.62/158.6 = 0.0102……. (mol)
n(H2O) = 1.07/18 = 0.0594……… (mol)
x = SrCl2 : H2O = 0.0594./0.0102……
= 5.8
dont always x by 10
A student repeats the titration to determine the molar mass and structure of A.
* The student prepares a 250.0cm3 solution from 1.513g of A.
* The solution of A is added to the burette and titrated with 25.0cm3 volumes of
0.112moldm–3 NaOH(aq).
* 1 mol of A reacts with 2 mol of NaOH.
* The student obtains a mean titre of 27.30 cm3.
(i) Calculate the molar mass of A from these results.
Give your answer to the nearest whole number. Show your working.
(NaOH)
= 0.112 25.0 = 0.00280 (mol) 1000
n(A) in 25.0 cm3
= 0.00280 = 0.00140 (mol)
2 n(A) in 250 cm3
= 0.00140 250.0 = 0.0128 (mol) 27.30
Molar mass, M(A) to nearest whole number.
= 1.513 = 118 (g mol–1) 0.0128
