Objective: To observe the single replacement reaction between iron and copper(II) sulfate, calculate the mole ratio and find percent error.
Theory: Mole relations: One mole (abbreviated mol) is equal to 6.02×10 23 molecular entities (Avogadro's number). Each element has a different molar mass depending on the weight of 6.02×10 23 of its atoms (1 mole). The total number of atoms of each element must be the same on each side of the equation to satisfy the Law of Conservation of Mass. The experiment aides in the understanding of mole-mass relationships that exist in a chemical reaction and in the interpretation of a balanced chemical equation.
Oxidation and reduction: Oxidation is the loss of electrons or an increase in oxidation state by a
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Quantitative Results:
Mass of empty beaker and label
50.95g
Mass of iron fillings
2.00g
Mass of beaker and dry copper
53.23g
Molar mass of copper
63.55g
Calculations:
Number of moles of iron used: nFe = m = 2.00g Fe =0.03581 = 3.58 x 10-2 mol M 55.85g/mol
Mass of copper produced: mCu= 53.23 - 50.95= 2.28 g
Number of moles of copper produced: nCu= m = 2.28 = 0.0358=3.58x10-2 mol M 63.55 Unreduced ratio mol of iron used : 3.58 x 10-2 mol Fe : 3.58 x 10-2 mol Cu mol of copper produced
Simplified mol ratio: 1 mol Fe :1 mol Cu
Percent Error: %Error= I actual - theoretical I x 100 = l2.28 - 2.28l x 100 = 0 % Theoretical
When the flame was blown out and the glowing wooden splint was placed halfway into the test tube containing H2O2 and MnO2 crystals, the splint reignited and caught flame once again. This demonstrates the decomposition of H2O2 into water and hydrogen. MnO2 is a catalyst that increases the rate at which H2O2 decomposes. Adding oxygen to a fire will cause it to burn faster and hotter and the oxygen rich test tube allowed the splint to reignite.
Cu (aq) + 2NO3 (aq) + 2Na+ (aq) + 2OH- (aq) → Cu(OH)2 (s) + 2Na+ (aq) + 2NO3(aq)
Measure the weight of a small stone to fit inside the opening of a 50ml graduated cylinder.
5) Find the experimental mass of both the c-clamp and the empty pan by using the formula from page one. Record this as mtotal.
* Find the density of both the steel and the honey. The equation Density = Mass Volume will be needed to calculate these. * Fill the measuring cylinder with 100cm3 of honey and place inside the water bath. * Fill the water bath with water of various different temperatures which are recorded before the experiment is carried out.
Atoms: 2 O-Oxygen Atomic weight →15.9994 Atoms: 1 Percent of mass: Na →74.1857% O →25.8143% NaOH mass: Na-Sodium Atomic weight → 22.98976928 Atoms: 1 O-Oxygen Atomic weight → 15.9994 Atoms: 1 H-Hydrogen Atomic weight→ 1.00794 Atoms: 1 Percent of mass: Na → 74.1857% O → 25.8143% H →
After the calculation we needed to convert pc to meters so we used the calculation factor of 1pc=3*10^16m
YEAR 7 SCIENCE Term 1 2016 Extended Experimental Investigation ‘Separating Mixtures’ REPORT BOOKLET Name:Harriet Slym Teacher:Mrs Laffranchi INTRODUCTION: In term one we learned about the separation of mixtures and in this assignment we are reporting on the separation techniques used to separate gravel, salt and iron. A mixture can be defined as many individual components combined together as one. A pure substance though is a material made up of only one particle like a Diamond, water, pure sugar, and gold.
There was 6.08% percent error in this experiment according to the data that was
The beginnings of modern processing of iron can be traced back to central Europe in the mid-14th century BC. Pure iron has limited use in today’s world. Commercial iron always contains small amounts of carbon and other impurities that change its physical properties, which are much improved by the further addition of carbon and other alloying elements. This helps to prevent oxidation, also known as rust.
11.) Subtract the mass of the evaporating dish from the mass of the evaporating dish and it's contents. Multiply that number by 10 to get the solubilty in grams per 100 cm3 of water.
get it exactly correct or you may not be able to measure the amount of
It was necessary to calculate the surface area of the aggregate to find out if adjustments needed to be made to the amount of binder that was required a simple formula was used to provide this information (Shell Bitumen 1991): T= b/(100-b) x 1/Db x 1/SAF
[6] Hayrynen KathyL., (2002) The Production of Austempered Ductile Iron (ADI), World Conference on ADI.