Answer:
u meant the answer to this ques
Explanation:
22
×12
------
44
22 ×
----------
264
-----------
or the answer can be 22×12=264
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
hey mate here is ur answer
solution
mass{m}=3 gram
=3/1000
volume{v}=16cm
=16/100
density=m/v
=3/1000÷16/100
=3/160
=0.01875kg/m3
Answer:
a) a0 was 46.2 grams
b) It will take 259 years
c) The fossil is 1845 years old
Explanation:
<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>
A = A0 * (1/2)^(t/h)
⇒ with A = the final amount = 46.2 grams
⇒ A0 = the original amount
⇒ t = time = 8 hours
⇒ h = half-life time = 3.2 hours
46.2 = Ao*(1/2)^(8/3.2)
Ao = 261.35 grams
<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>
t = (ln(0.66))-0.693) * 432 = 259 years
It will take 259 years
<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>
<em />
t = (ln(0.80))-0.693) * 5730 = 1845
The fossil is 1845 years old