Chemistry Questions (1 Viewer)
- Thread starter nsbrando
- Start date
someth1ng
Retired Nov '14
Re: Which one is oxidant and reductant
Are those ions? If they are ions, you MUST put it in.
someth1ng
Retired Nov '14
Re: name species undergoing oxidation
Cu undergoes oxidation to become CuCl2 because the oxidation state changes from 0 to +2 - that is, it loses electrons to the Cl2.
Zn undergoes oxidation to become Zn2+ because the oxidation state changes from 0 to +2 - that is, it loses electrons to Pb2+.
Cu undergoes oxidation to become CuCl2 because the oxidation state changes from 0 to +2 - that is, it loses electrons to the Cl2.
Zn undergoes oxidation to become Zn2+ because the oxidation state changes from 0 to +2 - that is, it loses electrons to Pb2+.
Re: Which one is oxidant and reductant
oxidant = causes oxidation. - therefore is the one that undergoes reduction. the ion that undergoes reduction (gain of e-) is the chlorine.
reductant = causes reduction=undergoes oxidation=loss of e- = iodine
oxidant = causes oxidation. - therefore is the one that undergoes reduction. the ion that undergoes reduction (gain of e-) is the chlorine.
reductant = causes reduction=undergoes oxidation=loss of e- = iodine
nsbrando
Member
- Joined
- Jun 19, 2012
- Messages
- 87
- Gender
- Male
- HSC
- 2013
half life multiple choice
1. Which sample will decay least over a period of 30 days?
a. 10g of I-131
b. 10g of P-32
c. 10g of Rn-222
d. 10g of Au-198
2. A sample of ^131 I decays to 1.0 gram in 40 days, What was the mass of the original sample?
a. 8.0g
b. 16g
c. 32g
d. 4.0g
3. What is the number of half-life periods required for a sample of a radioactive material to decay to one-sixteenth its original mass?
a. 8
b. 16
c. 3
d. 4
4. In how many days will a 12-gram sample of 131 / 53 I decay, leaving a total of 1.5 grams of the original isotope?
a. 24
b. 8.0
c. 16
d. 20
5. As a sample of the radioactive isotope ^131 I decays, its half-life
a. decreases
b. increases
c. remains the same
1. Which sample will decay least over a period of 30 days?
a. 10g of I-131
b. 10g of P-32
c. 10g of Rn-222
d. 10g of Au-198
2. A sample of ^131 I decays to 1.0 gram in 40 days, What was the mass of the original sample?
a. 8.0g
b. 16g
c. 32g
d. 4.0g
3. What is the number of half-life periods required for a sample of a radioactive material to decay to one-sixteenth its original mass?
a. 8
b. 16
c. 3
d. 4
4. In how many days will a 12-gram sample of 131 / 53 I decay, leaving a total of 1.5 grams of the original isotope?
a. 24
b. 8.0
c. 16
d. 20
5. As a sample of the radioactive isotope ^131 I decays, its half-life
a. decreases
b. increases
c. remains the same
Riproot
Addiction Psychiatrist
- Joined
- Nov 10, 2009
- Messages
- 8,228
- Gender
- Male
- HSC
- 2011
- Uni Grad
- 2017
1 - C
2 - Haven't done it yet.
3 - D
4 - See 2
5 - A?
2 - Haven't done it yet.
3 - D
4 - See 2
5 - A?
DamTameNaken
Member
- Joined
- Sep 27, 2012
- Messages
- 48
- Gender
- Male
- HSC
- 2013
Re: half life multiple choice
1. One with the longest half life (ignore mass for this one). You won't be expected to memorise the half lives in the HSC, and you should probably find the answer by googling it.
2. C 40 days = 5 half lives. i.e. original mass was 1 x 2^5. = 32 g
3. D. The formula is 1/(2^n) where n is the number of half lives
4. A. 1.5 is one eight of 12, which means there were 3 half live periods
5. C. Half life is always the same, in one period of half life you will always lose 50% of to radiation its just that remaining 50% gets smaller and smaller.
1. One with the longest half life (ignore mass for this one). You won't be expected to memorise the half lives in the HSC, and you should probably find the answer by googling it.
2. C 40 days = 5 half lives. i.e. original mass was 1 x 2^5. = 32 g
3. D. The formula is 1/(2^n) where n is the number of half lives
4. A. 1.5 is one eight of 12, which means there were 3 half live periods
5. C. Half life is always the same, in one period of half life you will always lose 50% of to radiation its just that remaining 50% gets smaller and smaller.
someth1ng
Retired Nov '14
Re: half life multiple choice
5 is C - the half life of an element is the half life of an element, it is a property of an element that won't change. It's like how molar mass of water will be the molar mass of water (18g/mol) but the mass of water depends on the sample.
1 is B - not really a HSC question but if it were to go into HSC, they'd give you 3 radioisotopes and 1 non-radioisotope. In this case, all 3 are radioactive and hence, you really need to search out each half-life to determine how much each decays.
5 is C - the half life of an element is the half life of an element, it is a property of an element that won't change. It's like how molar mass of water will be the molar mass of water (18g/mol) but the mass of water depends on the sample.
someth1ng
Retired Nov '14
Re: half life multiple choice
Yeah, 1 can also be used IF they give only 1 radioisotope and 3 stable nuclei.
nsbrando
Member
- Joined
- Jun 19, 2012
- Messages
- 87
- Gender
- Male
- HSC
- 2013
mola heat of combustion
The molar heat of combustion of 1-propanol is 2021 kJ mol-1. Assuming no heat loss, what would be the final temperature of the water???
Mass of 1 Propanol burnt: 0.60g
Mass of water heated: 200g
Initial temp of water: 21.0 degrees celcius
The molar heat of combustion of 1-propanol is 2021 kJ mol-1. Assuming no heat loss, what would be the final temperature of the water???
Mass of 1 Propanol burnt: 0.60g
Mass of water heated: 200g
Initial temp of water: 21.0 degrees celcius
Re: mola heat of combustion
change in H=m*C*change in temp
4.18*200*change in temp=836*change in temp
=0.836*change in temp
moles of fuel=grams of fuel/MM of fuel
moles of fuel=0.6/60.094
=0.001
Molar heat of combustion=2021 kJ/mol
and, we know molar heat of combustion is kJ/(mol of fuel) or enthalpy change/moles of fuel
therefore 2021 kJ/mol=(0.836*change in temp)/0.001 <--- (since we calculated previously enthalpy change in kJ was 0.836*change in temp whilst the moles of fuel are 0.001 (3 DP)
Rearranging the equation we get (2021*0.001)/(0.836)=change in temp
therefore change in temperature is 24.14 degrees Celsius. However that is the CHANGE in temp from intiial temp. so, final temp is 24.14+21=45.14 degrees Celsius
change in H=m*C*change in temp
4.18*200*change in temp=836*change in temp
=0.836*change in temp
moles of fuel=grams of fuel/MM of fuel
moles of fuel=0.6/60.094
=0.001
Molar heat of combustion=2021 kJ/mol
and, we know molar heat of combustion is kJ/(mol of fuel) or enthalpy change/moles of fuel
therefore 2021 kJ/mol=(0.836*change in temp)/0.001 <--- (since we calculated previously enthalpy change in kJ was 0.836*change in temp whilst the moles of fuel are 0.001 (3 DP)
Rearranging the equation we get (2021*0.001)/(0.836)=change in temp
therefore change in temperature is 24.14 degrees Celsius. However that is the CHANGE in temp from intiial temp. so, final temp is 24.14+21=45.14 degrees Celsius
Last edited:
someth1ng
Retired Nov '14
Re: mola heat of combustion
kunal96, don't forget your units during the calculation.
kunal96, don't forget your units during the calculation.