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Chemistry: Post your doubts here!

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I will try.

Your 1st question: The answer is "D", simply divide the number of moles of oxygen by the volume of bleach (in dm^3 unit), as the number of moles of oxygen correspond to the number of moles of NaClO, provided that hydrogen peroxide is in excess

2nd: To answer this, try to write out a balanced equation for the reaction to determine the ratio of the amount of reactant to that of product, but it is ok even if you choose not to write out. Just have to know that the aluminium will end up forming aluminium chloride (AlCl3) in a complete reaction. So 1 mol of aluminium will form 1 mol of AlCl3 with 3 mol of chlorine. For hydrochloric acid (HCl), 1 mol of HCl contains 1 mol of chlorine, so 3 mol of HCl is needed to form 1 mol of AlCl3 to completely react with 1 mol of aluminium.

In this case, 0.02 mol of aluminium is used, so 0.02 x 3 mol of HCl is needed, which is 0.06 mol of HCl. Here it says the concentration of HCl is 2.00 mol dm-3, so to provide 0.06 mol of HCl, the volume needed is (0.06 divided by 2.00) dm3. Convert the volume in dm3 to cm3 and you will get the answer which is "C", 30 cm3.

3rd: To answer this question, one way is to find out how many number of moles of carbon is contained in 35.2 gram of carbon dioxide and how many moles of hydrogen are there in 14.4 gram of water. Do this by dividing 35.2 by molar mass of carbon dioxide, whose quotient is about 0.765 mol, and dividing 14.4 by molar mass of water, whose quotient is about 0.8 mol. This is not yet over. 0.765 mol of carbon dioxide means 0.765 mol of carbon, whereas 0.8 mol of water means 1.6 mol of hydrogen atoms.

With the information so far, it is informed that 0.2 mol of hydrocarbon contains 0.765 mol of carbon and 1.6 mol of hydrogen atoms. The formula of hydrocarbon can be obtained by knowing how many carbon atoms and hydrogen atoms are in there in one molecule of hydrocarbon. No. of carbon atoms = 0.765 / 0.2 = 3.825 rounded off to 4. No. of hydrogen atoms = 1.6 / 0.2 = 8. So the formula of hydrocarbon is C4H8, a butane. The answer is D.

4th: First, know the mass of nickel in the coin, which is 10.0 g x 20/100 = 2g. Second, divide the mass of nickel by its molar mass to get its no. of moles, which is 2/58.7 = 0.0341 mol. Third, multiply the mol by Avogadro's Constant, which is 0.0341 x 6.02 x 10^23 = 2.05 x 10^22. The answer is A.

5th: If one Pb 4+ ion is reduced to Pb 2+ ion, it requires 2 Br - ions to do that. So 6.98 g of lead (IV) chloride contains 0.02 mol of lead (IV) chloride molecules (By dividing the mass by its molar mass). So it contains 0.02 mol of lead or therefore lead (IV) ions. So it needs 0.02 x 2 = 0.04 mol of bromide ions. 0.04 mol of bromide ions will be reduced to form 0.04 mol of bromine atoms. So 0.04 mol of bromine atoms has mass of 0.04 x its molar mass = 3.196 gram. The answer is C.

6th: It asks about the properties of chlorine atoms, which are always the same in whichever isotope of chlorine. So the isotopic mass is definitely not the answer because isotopic mass is different between different isotopes. As different isotopes of one element have different number of neutrons, so is the nucleon number, so this answer is out too. All is left is the atomic radius, which is the same for all isotopes of one element, because the number of protons and electrons is the same for every isotope, so is the structure of orbitals. This is just a simplified explanation, there are more reasons to account for the fact. So the answer is D.

7th: Simply determine how many moles of zinc are there in each statement, and how many moles of HCl in 0.1 dm3 of 1.00 mol dm-3 of the acid, and how many moles of hydrogen gas molecules (H2) in 9.6 dm3 of the gas ( the molar volume of gas varies in different temperature, but in this case try both 22.5 dm3 and 24.0 dm3 ). Then compare the number of moles of the reactant and that of product, using the balanced chemical equation. The answer is D, as only the first statement is correct.

8th: If 1 mol of monomers is fully polymerised, the most possible number of polymers is 0.5 mol if every polymer formed is a dimer, whereas the least possible number of polymers is 1/(6.02 x 10^23) mol if all monomers bond together into one (Note: 1 mol of particles consists of 6.02 x 10^23 of the particles). So the possible number of polymers formed is an integer lying in the range between the maximum and minimum numbers. So the answer is C, statement 2 and 3.

Hope they help, and correct me for any point mistaken.
 
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Can someone plz xplain this question:
A student mixed 25cm3 of 0.10mol dm–3 sodium hydroxide solution with 25cm3 of 0.10mol dm–3 hydrochloric acid and noted a temperature rise of 2.5°C.
What is the enthalpy change of the reaction per mole of NaOH?
A –209 kJ mol–1
B –104.5 kJ mol–1
C –209 J mol–1
D –522.5 J mol–1
 
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yar can u upload this document scribd does not give the access to download this


I've downloaded several files from scribd by uploading stuff (if you don't have anythign, uploading bogus files and taking them down)

Anyhow. From phone right now. WIll upload on dropbox or 4shared etc and send.
 
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675
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yar can u upload this document scribd does not give the access to download this


I've downloaded several files from scribd by uploading stuff (if you don't have anythign, uploading bogus files and taking them down)

Anyhow. From phone right now. WIll upload on dropbox or 4shared etc and send.
 
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today in my titration i got Ar of M in M2CO3 as 27.50 as M is group 1 metal i dont find any suitable metal with Ar of 27.50 should i write Na-23 as the answer as it is closest? or is my titration wrong. i got best titres 0.1cm3 apart so they seemed reliables. rough titre was 33.5cm, three titres were 32.00, 31.70, 31.80 so 31.75 seems correct to me. or maybe the laboratory guy mixed F1 and f3 in slightly wrong concentration. my working were quite good. My first time experience with As level experiment. what you think?
s2013/33
 
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today in my titration i got Ar of M in M2CO3 as 27.50 as M is group 1 metal i dont find any suitable metal with Ar of 27.50 should i write Na-23 as the answer as it is closest? or is my titration wrong. i got best titres 0.1cm3 apart so they seemed reliables. rough titre was 33.5cm, three titres were 32.00, 31.70, 31.80 so 31.75 seems correct to me. or maybe the laboratory guy mixed F1 and f3 in slightly wrong concentration. my working were quite good. My first time experience with As level experiment. what you think?
s2013/33

Close enough. I'd go with Na.
 
Messages
675
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today in my titration i got Ar of M in M2CO3 as 27.50 as M is group 1 metal i dont find any suitable metal with Ar of 27.50 should i write Na-23 as the answer as it is closest? or is my titration wrong. i got best titres 0.1cm3 apart so they seemed reliables. rough titre was 33.5cm, three titres were 32.00, 31.70, 31.80 so 31.75 seems correct to me. or maybe the laboratory guy mixed F1 and f3 in slightly wrong concentration. my working were quite good. My first time experience with As level experiment. what you think?
s2013/33

Close enough. I'd go with Na.
 
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Hi I'm new here,
Could someone help me please?
9701/1/O/N/02

35. the element astatine lies below iodine in Group VII of the periodic table.
What will be the properties of astatine?

1 It forms diatomic molecules which dissociate more readily than chlorine molecules
2 It reacts explosively with hydrogen
3 It is a good reducing agent

Answer is 1 only, why is 3 incorrect?

39. Which ions are present in a solution of ethanol in an excess of concentrated sulphuric acid?

1 CH3CH2O^-
2 CH3CH2O^+H2
3 HSO4^-

Why is the answer 2 and 3?

Help is very much appreciated! :)
 
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Hi I'm new here,
Could someone help me please?
9701/1/O/N/02

35. the element astatine lies below iodine in Group VII of the periodic table.
What will be the properties of astatine?

1 It forms diatomic molecules which dissociate more readily than chlorine molecules
2 It reacts explosively with hydrogen
3 It is a good reducing agent

Answer is 1 only, why is 3 incorrect?

39. Which ions are present in a solution of ethanol in an excess of concentrated sulphuric acid?

1 CH3CH2O^-
2 CH3CH2O^+H2
3 HSO4^-

Why is the answer 2 and 3?

Help is very much appreciated! :)

For Q35 - 3 is incorrect because it is a "good" oxidising agent (Oxidising abilites of the halogens decrease as you go down the group). Astatine wants 1 more electron for it to be more stable therefore it needs to get reduced (OIL RIG - Reduction Is Gain of electrons), hence it is a good oxidising AGENT being itself reduced.

For Q39- 2 is correct because: The solution of sulphuric acid dissociates in the soltuion -> H2SO4 ----> HSO4- + H+ , then the ethanol, CH2CH3OH can accept a H+ from the dissociation of the Sulphuric acid to form the + ethanol ion, CH2CH3O+H2. Hence you have the HSO4- ion and the CH3CH2O+H2 ion present in the solution.
Just in case, the ethanol gains a + charge because it gains a proton (H+) and the H2SO4 gains a - charge because it lost a proton (H+)

Hope it helped.
 
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Q20
M/J 12 12
when this ester is hydrolysed, 2 products r formed. one is propanol and the other is sodium ethanoate.
Mr of propanol 60
Mr of sodium ethanoate 82
calculate %age
propanol 60/142*100= 42.3
Na ethanoate 82/142*100= 57.7:)
 
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Q 23
M/J 12 12
it is the addition of HBr to both cis trans isomers.
Answer is B.
in A , the 3rd compound is wrong. in it H2 is added rather than HBr.
in C, in 1st compound, Br2 is added.
in D, the first compound. in it carbon is forming 5 bonds. which cant happen
in B all compounds r corrct
 
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