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Simora [160]
3 years ago
11

41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay re

action? 41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay reaction? Ti Cl Sc Ar Ca
Chemistry
1 answer:
dimaraw [331]3 years ago
4 0

Answer:

Cl

Explanation:

⁴¹Ca has an atomic number equal to 20, it means that it has 20 protons and 20 electrons ad its neutral state. In the decay by electron capture, it will lose one electron and will become a cation, but the mass (41) and the atomic number will remain the same.

When Ca⁺ undergoes alpha decay, it will lose an alpha particle, which has mass 4 and 2 protons.

⁴¹₂₀Ca⁺ → ³⁷₁₈X⁺ + ⁴₂α

To be stable, X will lose a proton and will become ³⁷₁₇X. The element which has atomic number 17 is chlorine, Cl.

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When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced
Vika [28.1K]

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO_2 and H_2O

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

When only 2.5 mol of O_2 are consumed in order to complete the reaction, ________ mol of CO_2 are produced.

Answer: Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

Explanation:

The balanced chemical equation is:

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

According to stoichiometry :

5 moles of O_2 produce = 3 moles of  CO_2

Thus 2.5 moles of O_2 will produce = \frac{3}{5}\times 2.5=1.5 moles of  CO_2

Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

4 0
3 years ago
Write a paragraph to explain the answer to the following prompt:
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Answer:

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The law of reflection states that the angle of incidence and the angle of reflection are always
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Answer:

The law of reflection states that the angle of incidence and the angle of reflection are always equal.

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Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) => CH 3OH(l) Using the following information: C(graphite) +
Alika [10]

Answer:

\Delta H for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

C(graphite)+O_{2}(g)\rightarrow CO_{2}(g) \Delta H_{1}= -393.5 kJ

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CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)  \Delta H_{3}= 726.4 kJ

---------------------------------------------------------------------------------------------------

C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)

\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ

4 0
4 years ago
The diagram represents liquid water in a pan on a hot plate. The liquid water is boiling and changing into water vapor. The proc
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Answer:

Physical change

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