# Can somebody tell me the answers to these questions they came up in my test today?Watch

#1
1) Calculate the volume of hydrogen gas, measured at room temperature and pressure, which would be formed when 12 g of magnesium react with excess dilute sulphuric acid (assume that one mole of any gas occupies a volume of 24,000 cm^3 (24dm^3) at room temperature and pressure; relative atomic mass of mg = 24). (2)
Mg(s) + H2SO4(aq)--->MgSO4(aq) + H2(g)

2) A car engine burns the compound octane C8H18
2C8H18(g) + 25O2(g)---> 16CO2(g) + 18H2O(l) Calculate:

(a) The volume of oxygen required to completely burn 2.00 dm^3 of octane vapour. (1)

(b) The volume of carbon dioxide gas you would expect to be formed. (1) (Assume that all gas volumes are measured at the same temperature and pressure.)

3) In a titration, it was found that 28.5cm^3 of dilute nitric acid (concentration 0.100 mol dm^-3) were exactly neutralised by 25.0 cm^3 of pottasium hydroxide solution.
KOH(aq) + HNO3(aq)--->KNO3(aq) + H2O(l)
Calculate the concentration of the potassium hydroxide solution in mol dm^-3. (2)

4) 40.0 cm^3 of an aqueous solution of pottasium hydroxide containing 5.6 g dm^3 are neutralised exactly by 40.0 cm^3 of dilute hydrochloric acid:
KOH(aq) + HCL(aq)--->KCL(aq) + H2O(l)
Calculate the concentration of the hydrochloric acid solution in mol dm^-3 (relative atomic masses K= 39, O= 16, H=1) (3)

5) In an organic chemistry experiment, 49.2 g of nitrobenzene (C6H5NO2) were obtained from 39.0 g of benzene (C6H6). The equation for the reaction is:
C6H6(l) + HNO3(l)---> C6H5NO2(l) + H2O(l)
Calculate the percentage yield of the reaction (relative atomic mass C= 12, H= 1, N= 14, O=16). (2)

really hard test tbh
0
7 years ago
#2
what do you know so far about mole? state all the formulas about moles that you know of. state what topics about mole concepts have you learnt so far.

then people can decide whether you are worth the "whole answer" help or just some "nudge in the right direction".
0
#3
topic ive covered
1) Counting chemical substances in bulk
2)Calculation of empirical and molecular formula
3) Balancing chemical equations
4) Balancing ionic equations
5) Calculations involving concentrations and gas volumes

I know that avoogadros numbers is 6.022 x 10^23
i know how to balance equations
number of moles= volume x concentration
volume= number of moles/concentration and
Concentration= number of moles/volume
i know how to work out the ratio for empirical formulas
0
7 years ago
#4
mole = volume x concentration(yes in solution)

how about mole for solids or gases?

those don't have concentrations.
0
7 years ago
#5
(Original post by trrr)
1) Calculate the volume of hydrogen gas, measured at room temperature and pressure, which would be formed when 12 g of magnesium react with excess dilute sulphuric acid (assume that one mole of any gas occupies a volume of 24,000 cm^3 (24dm^3) at room temperature and pressure; relative atomic mass of mg = 24). (2)
Mg(s) + H2SO4(aq)--->MgSO4(aq) + H2(g)

2) A car engine burns the compound octane C8H18
2C8H18(g) + 25O2(g)---> 16CO2(g) + 18H2O(l) Calculate:

(a) The volume of oxygen required to completely burn 2.00 dm^3 of octane vapour. (1)

(b) The volume of carbon dioxide gas you would expect to be formed. (1) (Assume that all gas volumes are measured at the same temperature and pressure.)

3) In a titration, it was found that 28.5cm^3 of dilute nitric acid (concentration 0.100 mol dm^-3) were exactly neutralised by 25.0 cm^3 of pottasium hydroxide solution.
KOH(aq) + HNO3(aq)--->KNO3(aq) + H2O(l)
Calculate the concentration of the potassium hydroxide solution in mol dm^-3. (2)

4) 40.0 cm^3 of an aqueous solution of pottasium hydroxide containing 5.6 g dm^3 are neutralised exactly by 40.0 cm^3 of dilute hydrochloric acid:
KOH(aq) + HCL(aq)--->KCL(aq) + H2O(l)
Calculate the concentration of the hydrochloric acid solution in mol dm^-3 (relative atomic masses K= 39, O= 16, H=1) (3)

5) In an organic chemistry experiment, 49.2 g of nitrobenzene (C6H5NO2) were obtained from 39.0 g of benzene (C6H6). The equation for the reaction is:
C6H6(l) + HNO3(l)---> C6H5NO2(l) + H2O(l)
Calculate the percentage yield of the reaction (relative atomic mass C= 12, H= 1, N= 14, O=16). (2)

really hard test tbh
1) 12dm^3
2)a) 25/24 dm^3
b) 16dm^3
3) 0.114moldm^-3
4)0.1moldm^-3
5) 80%
Just solved the questions roughly myself so can't confirm their accuracy but they will probably be correct.
NB: Scored an A*, 572/600, in chemistry a-level last year .
4
7 years ago
#6
(Original post by raheem94)
1) 12dm^3
2)a) 25/24 dm^3
b) 16dm^3
3) 0.114moldm^-3
4)0.1moldm^-3
5) 80%
Just solved the questions roughly myself so can't confirm their accuracy but they will probably be correct.
NB: Scored an A*, 572/600, in chemistry a-level last year .
man, i was hoping i could give him some hints, instead of answers.
1
#7
i also know how to balance ionic equations and im not sure about what u mean about the mole of solids and gases.
One mole of a substance is the amount of substance that has the same number of particles as there are atoms in exactly 12 g of carbon- 12. This number is called Avogadros number.

empirical formula shows the simplest form of whole number rations(similar to the canceling down of fractions). The molecular formula shows the ratio at its original form.

Molar masses enable calculations to be made using moles and balanced chemical equations. such calculations involve reacting masses, volume and concentrations of solutions, or volume of gases *yawn*
0
#8
(Original post by shengoc)
man, i was hoping i could give him some hints, instead of answers.
dont worry i havent looked at the answerrs just yet just i really wanna know how it has been worked out and to see how it compares to the way i worked it out
0
7 years ago
#9
(Original post by trrr)
1) Calculate the volume of hydrogen gas, measured at room temperature and pressure, which would be formed when 12 g of magnesium react with excess dilute sulphuric acid (assume that one mole of any gas occupies a volume of 24,000 cm^3 (24dm^3) at room temperature and pressure; relative atomic mass of mg = 24). (2)
Mg(s) + H2SO4(aq)--->MgSO4(aq) + H2(g)

mol of solids = mass in g/molecular mass in g per mol

mol of gas = volume in dm cube/ 24 dm cube per mol = volume in cm cube/24 000 cm cube per mol

NB. 1 dm cube = 1 litre = 1000 ml = 1000 cm cube

2) A car engine burns the compound octane C8H18
2C8H18(g) + 25O2(g)---> 16CO2(g) + 18H2O(l) Calculate:

(a) The volume of oxygen required to completely burn 2.00 dm^3 of octane vapour. (1)

(b) The volume of carbon dioxide gas you would expect to be formed. (1) (Assume that all gas volumes are measured at the same temperature and pressure.)

3) In a titration, it was found that 28.5cm^3 of dilute nitric acid (concentration 0.100 mol dm^-3) were exactly neutralised by 25.0 cm^3 of pottasium hydroxide solution.
KOH(aq) + HNO3(aq)--->KNO3(aq) + H2O(l)
Calculate the concentration of the potassium hydroxide solution in mol dm^-3. (2)

(a) construct balanced chemical eqn involving nitric acid and potassium hydroxide.
(b) balance the equation.
(c) mole = concentration in mol per dm cube x volume in dm cube

SOLUTION:
a) equation is balanced.
b) mol of acid = 28.5/1000 x 0.100 = ? (work this out)
c) mol of KOH = y x 25/1000 (y is the unknown concentration of KPH)
d) from balanced equation,
mol of KOH/mol of HNO3 = 1/1
do cross multiply, then
1 x mol of KOH = 1 x mol of HNO3
hence ? (above) = y x 25/1000
solve for y, the unknown concentration

4) 40.0 cm^3 of an aqueous solution of pottasium hydroxide containing 5.6 g dm^3 are neutralised exactly by 40.0 cm^3 of dilute hydrochloric acid:
KOH(aq) + HCL(aq)--->KCL(aq) + H2O(l)
Calculate the concentration of the hydrochloric acid solution in mol dm^-3 (relative atomic masses K= 39, O= 16, H=1) (3)
like above, ratio is very important, get it right everytime.

5) In an organic chemistry experiment, 49.2 g of nitrobenzene (C6H5NO2) were obtained from 39.0 g of benzene (C6H6). The equation for the reaction is:
C6H6(l) + HNO3(l)---> C6H5NO2(l) + H2O(l)
Calculate the percentage yield of the reaction (relative atomic mass C= 12, H= 1, N= 14, O=16). (2)

really hard test tbh
left some hints for you
0
7 years ago
#10
chemical reactions in lab often don't give 100% yield, meaning if you expect 10 g to come out, not all 10 g would come out as products. we say such reaction has a percentage yield of however much it is, often less than 100%.

so % yield = mass you actually get/mass you expect to get from 100 percent reaction x 100%

in this case, for question (5), which is your experimental mass(the mass of product you actually get)?

use previous methods to find out what mass you expect to get.
0
7 years ago
#11
(Original post by shengoc)
man, i was hoping i could give him some hints, instead of answers.
The OP looks new to chemistry at a-level so he needed answers to correct his mistakes. I would be wrong if i had posted the solutions but these are just the answers.
0
7 years ago
#12
(Original post by shengoc)
man, i was hoping i could give him some hints, instead of answers.
if hes done the test already then it doesnt matter if you give the answers
0
7 years ago
#13
(Original post by s04582)
if hes done the test already then it doesnt matter if you give the answers
still, it'd be nice if he actually learns how to do the questions based on what he knows at the moment rather than just being told what is what. perhaps that is what schools are becoming more and more like these days.
0
#14
(Original post by shengoc)
still, it'd be nice if he actually learns how to do the questions based on what he knows at the moment rather than just being told what is what. perhaps that is what schools are becoming more and more like these days.
so all day i had to do is divide 24/ by the ration for solids and about the same thing for liquids
0
7 years ago
#15
(Original post by trrr)
so all day i had to do is divide 24/ by the ration for solids and about the same thing for liquids
why 24? do you even know molar volume for gas at rtp is 24 mol per dm cube.

like i said, i dont believe your textbook or class notes dont include mole formula for solids or solutions?
0
#16
(Original post by shengoc)
why 24? do you even know molar volume for gas at rtp is 24 mol per dm cube.

like i said, i dont believe your textbook or class notes dont include mole formula for solids or solutions?
what class notes lol, i suggested 24 becuase the mass of mg is 24. plus this isnt all the question in my test there was about ten othes but these were hardest
0
#17
and yes i do know that gases at rtp is 24dm^3
0
#18
well either 24 or 12 anyway
0
7 years ago
#19
(Original post by trrr)
what class notes lol, i suggested 24 becuase the mass of mg is 24. plus this isnt all the question in my test there was about ten othes but these were hardest
well the situation is like i could mention any number like 6. you won't know exactly what it means, but hey, i was trying to mean that it is lithium mass number. you get my drift? be specific, be clear, be concise.
0
#20
(Original post by shengoc)
well the situation is like i could mention any number like 6. you won't know exactly what it means, but hey, i was trying to mean that it is lithium mass number. you get my drift? be specific, be clear, be concise.
yh, i doubt i passed dis test anyway, i need a tutor asap
0
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