Hept for 7, hence the number is seven.
Answer:
53.85%
Explanation:
Data obtained from the question include:
Mass of antimony (Sb) = 27.6g
Mass of Fluorine (F) = 32.2g
Mass of compound = 59.8g
Percentage composition of fluorine (F) =..?
The percentage composition of fluorine can be obtained as follow:
Percentage composition of fluorine = mass of fluorine/mass of compound x 100
Percentage composition of fluorine = 32.2/59.8 x 100
= 53.85%
Therefore, the percentage composition of fluorine in the compound is 53.85%
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g
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