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tigry1 [53]
3 years ago
7

How much of a sample of radio-isotope remais after one half life?

Chemistry
1 answer:
Tatiana [17]3 years ago
3 0

Half-life is defined as the time which is taken to reduce the initial amount of isotopes by half.


Hence, after one half-life, half of initial amount of isotopes will remain.

<span>
</span>

<span>The equation for the half-life, t1/2 = ln 2 / </span>λ,<span> where </span>λ <span>is the decay constant. </span>

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Which of the following statements are true about whether atoms tend to gain<br> or lose electrons?
Dafna11 [192]

Answer:

C and D

Explanation:

Atoms with five, six or seven valance electrons gain electrons to complete the octet because it is more convenient for the atoms to gain three, two or one electron as compared to lose five, six or seven electrons. Thus atoms with five, six or seven valance electrons form negative ions by gaining electrons.

Atoms with one, two or three valance electrons lose the electrons to get complete octet because it is more convenient for the atoms to lose one two or three electrons as compared to gain the seven, six or five electrons. The atoms with one, two or three valance electrons form positive ions.

5 0
3 years ago
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Brut [27]

Answer:

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Explanation:

if it's the white ring I think your talking abt

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What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?
Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

6 0
2 years ago
6. The modern view of the atom has come a long way from that of a solid, indestructible sphere
Fynjy0 [20]
It is a true statement
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Write the balanced chemical equation for the reaction of glucose (C6,H12,O6) with oxygen gas to produce carbon dioxide gas
Fittoniya [83]

Answer:

C6H12O6+6O2=6Co2+6H2o

4 0
3 years ago
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