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None of the compounds listed or any possible combination there of should turn red.

m1 = 0.728 g m2 = 0.773 g m3 = 0.722 g m4 = 0.732 g ma = 0.73875 g 0.73875 g * 10 = 7.3875 g 7.3875 g / 50 g * 100 % = 14.775 % --- Adding water to the 50 mL sample does not change the amount of sulfate. An aliquot of one tenth the volume is removed so logically one tenth of the sulfate is...

Phosphorous is a non-metal. Elemental metals are anything that exhibit metallic bonding and are elementally pure.

Say you wanted to liquify nitrogen and oxygen from the air. Dry air is composed of nitrogen, oxygen, argon and carbon dioxide plus traces of other gases. Nitrogen condenses at 77.36 K, oxygen at 90.20 K, argon at 87.30 K and carbon dioxide at 216.6 K. The order of condensation will be carbon...

HOBr is hypobromous acid not bromine hydroxide. The ionisation products are H+ and BrO-. The products of Br2(l) + H2O(l) --> HOBr(aq) + HBr(aq) are both acids.

Br2 + H2O --> BrOH + HBr

It should be part of the appendix at the end of the report followed by all the data.

The 1,2- indicates that there is a bromine atom on carbon 1 and one on carbon 2.

You will need to greatly expand on the dot point answers if the assessment is a report.

Br2(l) + H2O(l) --> Br2(aq) Br2(l) + H2O(l) --> HBr(aq) + HOBr(aq)

Youve drawn a substitution reaction. Your product is still an alkene. Bromine + alkene is an addition reaction to form an alkane. C6H12 + Br2 --> C6H12Br2 Although technically the reaction is: Br2 + H2O --> HBr + HOBr and both C6H12Br2 and C6H12BrOH are formed.

It is coloured. When it reacts with the alkene it forms a colourless product. The change from coloured to colourless indicates that a reaction has occured. Whereas if you were going to add water across the double bond, the mixture would start colourless and remain colourless. You cant be certain...

C D A H = m * C * dT H = 100 * 4.18 * 17 H = 7106 kJ.mol-1

It will precipitate during flocculation as aluminium phosphate or calcium phosphate depending on the flocculant.

No Both of those equations are electrically unbalanced. CFCl3 --> CFCl2. + Cl. CFCl3 --> CFCl2+ + Cl- See the difference? One of the C-Cl bonds has broken. In the first equation the C keeps one electron and the Cl keeps the other electron. They are both radicals. In the second equation the C...

The . represents a single electron which is required by a radical. The - represents a pair of electrons which is required by an anion. So yes the . must be displayed.

A couple more points for you all Octane is immiscible in water and less dense so in the event of a spill it will form a distinct layer on top of water allowing for easy clean up. Where as ethanol is miscible in water so the result will be a toxic alcohol solution which cannot be cleaned...

Methane boils at -161.5 oC, ethane at -88.6 oC, and ethene at -103.7 oC. There wont be a time when any of them are liquid.

Do not ever round off in the middle of a calculation. Give the answer with all of the decimal places and then again to the required number. If the answer is being used in another calculation use all the decimal places from the first answer.

People say that every year and one year soon theyre going to get bitten on the arse. Learn everything in the syllabus.