me again!
got a question that really is trying!
the isotope of ununhexium Uuh and it undergoes a succession of seven a-decay reactions. I need to find out what the final isotope would be as a result. And is the final product a lanthanide, an actinide, a transition element or a typical element?
Any help would be great ;)
I've asked the same question, are you studying with the OU?
If so you may be able to help me with the questions I'm stuck on, please!
iam indeed! S103!
wot one u on?
You just must know how to add to do this problem. An alpha particle is a helium nucleus. That means it is 2He4.
Uuh is mass number 292. atomic number 116.
116Uuh292 ==> 7 2He4 + 102X264
Note that the atomic masses added up on each side and the mass numbers add up on both sides. Now all you need to do is to determine what X stands for. Look at your periodic table and find element #102 and replace X with that symbol. And you have it. simple, Huh?
If I didn't get all the subcripts and superscripts right I will correct it soon.
They look ok. Note that 116 on the left = 7*2 + 102 on the right.
And 292 on the left = 7*4 + 264 on the right. Ain't science wonderful?
yes the same, are you doing tma 5?
its hard i kno that much!!
Im not getting how u reached that though. I can see where u got the masses and that from, but how did you know that's what Uuh was made up of helium and nobelium? Am i missing something obvious?
Sorry!
afraid so!
do you know the answe to 2a(ii) and 2b (iii), (iv) and (v)
I've been at this for twelve hours today and i just can't figure it out!
im the exact same!! its took about four days to get this far, its soooo annoying!
im worried that we mite not be allowed on this site, wot do u think?
Im not sure my answers are correct tho that's the thing!
My head was fried but the answers are in the book at the answer section, but ill help as much as i can!
wot region are u in?
Yes, you are missing something. Uuh is an element, #116. I know that because when you say Uuh, I go to the periodic table, find it, and it is atomic number 116 and the table tells me it has a mass of 292 (at least one of the isotopes of Uuh has a mass of 292). BUT the nucleus is not stable. In fact, the nucleus of none of the elements above U are stable (and some below U are not stable). This is radioactivity. Therefore, the nucleus tries to regain stability by emitting alpha particles, beta particles, and/or gamma rays. The problem TELLS you that Uuh emits 7 alpha particles and all you need to know is that an alpha particle is a He nucleus with an atomic number of 2 and a mass number of 4. The rest of it is simple subtraction to make the right side equal to the left side. In rough form, this is what I went through.
116Uuh292 ==> 7*2He^4 + ??X??
On the right, protons are 7*2=14 so 116-14=102 so the first ?? is 102. The second set of ??. 4*7=28 and 292-28=264.
Then I look up 102 and that is No and I'm in business.
I'm in iow, don't think they could tell exactly who you are. loads of people use it anyway. Where are you?
what answers did you get?
if there are any i can help you with, i will if i can
aaaah, i get you now! the 7 alpha particles is wot i missed, oops!
Thanks for being soooo patient! ;)
is that ur region? where abouts are u?
Im in Scotland!
Do u just want the answer or the whole lot?
your rite, i doubt they could tell tho! i should NEVER have picked this course, its torture!!
Isle of Wight,
the whole lot please
thank you so much x
ii.
you need to use the volume you worked out in i.
i got 176cm3 (prob wrong, but i don't care anymore hehe)
pressure = 760mmHg
Temp = 650c
New temp = 0c
oc = 232k nad 650c = 923k
volume at 273k = 176cm3/923k = 52cm3
it then asks for litres, so ive just done:
1 litre = 1000cm3
52cm3/1000cm3
0.052l
5.2 x 10 to the power of -2
Does that make sense?
i should have guessed really!
Did u do q 1 di and ii?
is it the set up shown on page on page 18?
yes i got 176 aswell
I don't think so, I thought that originally, but I saw a question about it on here this morning and the answer made sense.
Hang on, i'll try and find it
phew, thank god haha!
Density = mass/volume
mass = 0.927g
volume= 52cm3
0.927g/52cm3 = 0.0178 (can u remember the unit for density?) oops!
density for litres = 0.0178/1000 = 0.0000178
1.78 x 10 to the power of -5 litres
Im not sure if this is correct or not, but im going with it!!
iv.
oxygen = 24% of oxide of arsenic
arsenic = 76%
24% of 0.927g = 0.927/100x24 = 0.222g
76% of 0.927 = 0.927/100x76 = 0.704g
As mols = 0.704g/As atomic mass 74.9 = 0.0093 mols
o mols = 0.222g/16.0 = 0.0138 mols
0.0093 - 0.0093 = 1
0.0138 - 0.0093 = 1.5
multiply 1 by 2 to give a whole number and multiply 1.5 by 2 to give a whole number
ratio = 2:3 therefore As3O2
Sound ok?
and finally v.
Im not sure of this one to be honest.
It leaves us with an odd number of electrons that cant be balanced.
like i say tho, i don't kno if its correct or not!! hope u don't fail now hehe X
answer to d(i)and (ii)
sorry took a while to find again
In this piece of homework I am asked to describe the process I would use to prepare a sample of dry hydrogen selenide.
I am given serveral materials to do so :
Liquids :
concentrated nitric ccid
concentrated hydrochloric acid
water
Solids :
calcium chloride granules
sodium chloride
aluminium selenide
sodium hydroxide
Solid aluminium selenide will react with water to form solid aluminium hydroxide [Al(OH)3] and hydrogen selenide gas (H2Se). I need to isolate and dry the hydrogen selenide gas (which is denser than air).
How would I do this?
EDIT : After some thought, I've considered this :
I'll use two addition funnels, the top with water in it, and the bottom with aluminium selenide. Water will be allowed through by undoing the valve, and the two substances will react. The valve on the water funnel will then be closed and the gas will escape through a tube on the side. The hydrogen selenide gas will then pass through a u-tube filled with calcium chloride granules to dry the gas. Finally, another tube will take the gas into a jar for collection .
Would this work?
Thanks.
mmm yeah i can c that! just wish someone had answered it tho! is that wot uve done?
For your information, that As3O2 should be As2O3.
yes that's the answer that i've used.
thanks for your help
Oooh! That's so naughty - telling people the wrong answers just to make them fail.