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3 years ago

What percentage of a radioactive species would be found as daughter material after six half–lives?

1 answer:

3 0

100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................

After six half-lives would be found 1.5625% of readioactive species.

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ANSWER QUICK

Novosadov [1.4K]

Answer:

1.8

Explanation:

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2 years ago

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If you know the atomic number and mass number of an atom of an element, how can you determine the number of protons, neutrons, a

ser-zykov [4K]

Answer:

You subtract the atomic number from the mass number to find the number of neutrons. If the atom is neutral, the number of electrons will be equal to the number of protons.

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3 years ago

What is the Golden Rule of Solubility? A. polar dissolves nonpolar. B. nonpolar dissolves polar. C. like dissolves unlike D. lik

DerKrebs [107]

Answer: D. like dissolves like

Explanation:

The solubility of substances is governed by: Like dissolves like, which states that polar compounds are soluble in polar solvents and non polar compounds are soluble in non polar solvents.

Hydrocarbons are non polar in nature due to less difference between the electronegativities of carbon and hydrogen and thus are soluble in non polar solvents only.

Ionic compounds which are formed by elements with high electronegativity difference are polar in nature and thus dissolve in polar solvents.

Example: $NaCl$ in water.

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3 years ago

Which geologic features help scientists determine the relative ages of rocks by their positions? Select three options.

stiks02 [169]

Answer:

Option D, Index fossil helps scientists to determine the relative ages of rocks by their positions

Explanation:

Index fossils are arranged in layer in such a way that the lowest layer represent the oldest fossil and the top most layer represents the youngest fossil. Scientist use this concept to determine the relative age of the rocks based on their position beneath the earth’s surface

Hence, option D is correct

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2 years ago

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g

Tanya [424]

Answer:

(A) $\Delta H^{\circ }_{r}= -144 kJ$

(B) $\Delta H^{\circ }_{r}= - 2552kJ$

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

$1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)$

$1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)$

Now, multiplying equation (a) with 2:

⇒ $N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)$

Then equation b is reversed and multiplied with 2:

$2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)$

Now by adding the equation (a) and equation (b), we get:

⇒ $2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)$

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

$\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}$

$= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ$

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

$B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)$

$2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)$

$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)$

$H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)$

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

$6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)$

$4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)$

$6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)$

$6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)$

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

$\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }$

$= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ$

4 0

3 years ago