<h3>
Answer:</h3>
6.25 atoms
<h3>
Explanation:</h3>
<u>We are given</u>;
- The half life of Po-218 is 3 minutes
- Initial sample is 200 atom
- Time of decay is 15 minutes
We are required to calculate the remaining mass after decay;
Half life refers to the time taken for original amount of a radioactive sample to decay to a half.
To calculate the remaining mass we use the formula;
N = N₀ × 0.5^n where n is the number of half lives, N is the new amount and N₀ is the original amount.
n = 15 min ÷ 3 min
= 5
Therefore;
New amount = 200 atom × 0.5^5
= 6.25 atoms
Therefore; the amount of the sample that will remain after 15 minutes is 6.25 atoms.
Answer:
A mixture!
Explanation:
Mixtures are a substance made of combining two or more different kinds of matter :)
Hope this helped!
Mass of BaO in initial mixture = 3.50g
Explanation:
Let mass of BaO in mixture be x g
mass of MgO in mixture be (6.35 - x) g
Initially CO_2
Volume = 3.50 L
Temp = 303 K
Pressure = 750 torr = 750 / 760 atm
Applying ideal gas equation
PV = nRT
n = PV / RT
(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303
(n)_CO_2 = 0.139 mole
Finally; mole of CO_2
n= PV /RT
((245/760) *3.5) / 303* 0.0821
(n)_CO_2 = 0.045 mole
Mole of CO_2 reacted = 0.139 - 0.045
=0.044 mole
BaO + CO_2 BaCO_3
Mgo + CO_2 MgCO_3
moles of CO_2 reacted = ( moles of BaO + moles of MgO)
moles of BaO in mixture = x / 153 mole
moles of MgO in mixture = 6.35 - x mole / 40
Equating,
x/ 153 +6.35/40 = 0.094
= x/153 + 6.35 / 40 - x/40 =0.094
= x (1/40 - 1153) = (6.35/40 - 0.094)
= x * 10.018464
= 0.06475
mass of BaO in mixture = 3.50g
Answer:
Explanation:
Hello,
In this case, the first step is to compute the number of moles of potassium phosphate in 20.0 mL (0.020L) of the 0.015-M (mol/L) solution as shown below:
Thus, these moles correspond to potassium phosphate moles, which molecular formula is K₃PO₄, therefore, one mole of this compound contains three moles of potassium ions as it has three as its subscript in the formula. Thereby, the moles of potassium ions result in:
Best regards.
The empirical formula of the compound is C₄H₅O₃.
<u>Explanation</u>:
If hydrogen % = 5.98823, the Sum of C+O = 94.01177% or 47.005885% each
The original mass of 300 g is not important
C = 47.005885 / 12.011 = 3.913569
H = 5.98823 / 1.008 = 5.94070
O = 47.005885 / 15.999 = 2.93805
Divide by the smallest number:
C =1.33
H = 2.022
O = 1
Multiply through by 3
C = 4
H =6
O = 3
The empirical formula of a compound = C₄H₆O₃
Do not be confused by the 300 g. It is totally irrelevant to the question because you are dealing with % amounts. It would only have been of importance if you were given some mass values.
