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2 years ago

If the passing of five half-lives leaves 25.0 mg of a strontium-90 sample, how much was present in the beginning

Chemistry

1 answer:

hodyreva [135]2 years ago

8 0

Answer:

initial amount = 800 mg

Explanation:

Given data:

Amount of strontium-90 after 5 half life = 25 mg

Total amount of sample = ?

Solution:

At 5th half life = 25 mg

At 4th half life = 25× 2= 50 mg

At 3rd half life = 50×2 = 100 mg

At 2nd half life = 100×2 = 200 mg

At 1st half life = 200×2 = 400 mg

when time 0 = 400×2 = 800 mg

Conformation:

initial amount = 800 mg

1st half life = 800/2 = 400 mg

2nd half life = 400/2 = 200 mg

3rd half life = 200/2 = 100 mg

4th half life = 100/2 = 50 mg

5th half life = 50 /2 = 25 mg

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Sever21 [200]

The given dehydration equation is,

$Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)$

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.

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2 years ago

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Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

5 0

2 years ago

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Explain why liquids unlike gases are virtually incompressible

MaRussiya [10]

Lipids cannot be compressed since there is only a small distance between the molecules when bonded

8 0

2 years ago

A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou

Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0

2 years ago

Chlorine can bond with fluorine to form CIF. Chlorine can also bond with lithium to form LiCI Which compound will have a greater

Sergio [31]

Answer: $ClF$ will have a greater partial charge.

Explanation:

A polar covalent bond is defined as the bond which is formed when there is a low difference of electronegativities between the atoms, thus resulting in charge difference. Example: $ClF$

Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms and thus there is no charge difference. Example: $F_2$

Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal. The electronegative difference between the elements is high. The charges on cation and anion neutralise each other. Example: $LiCl$

Thus as $ClF$ will have greater partial charge.

5 0

2 years ago