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tester [92]
3 years ago
12

Determine the identity of the daughter nuclide from the electron capture by 3517cl. determine the identity of the daughter nucli

de from the electron capture by . 4019k 3218ar 3516s 3215p 3718ar
Chemistry
1 answer:
brilliants [131]3 years ago
4 0

Answer:

  •           ^{35}_{16}S

Explanation:

When the nucleus of an <em>atom captures and electron</em>, such electron combines with a proton and forms a neutron. Thus, the mass number remains the same (just a proton has been converted into a neutron) but the atomic number (the number of protons) decrease in one.

Then, the daughter nuclide will have the same mass number and the atomic number reduced in one.

The given parent isotope is    ^{35}_{17}Cl,    which means that it has these features:

  • Chemical symbol: Cl
  • Name of the element: chlorine
  •  Mass number (the superscript to the left): 35
  •  Atomic number (the subscript to the left): 17

And the daughter nuclide after the electron capture will be:

  • Mass number: 35 (the same of the parent nuclide)
  • Atomic number: 17 - 1 = 16
  • Identity: sulfur (search the element with atomic number 16 in a periodic table).
  • Chemical symbol: S

  • Idenity of the daughter nuclide using the isotope notation:

          ^{35}_{16}S

Hence, the answer is the third choice.

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hope that helps you please mark brainliest
3 0
3 years ago
Examine the given reaction. NH4NO3(s) → NH4+(aq) + NO3–(aq) ΔH° = 25.45 kJ/mol ΔS° = 108.7 J/mol·K Which of the given is correct
kobusy [5.1K]

Answer:

B)−6,942 J /mol

Explanation:

At constant temperature and pressure, you cand define the change in Gibbs free energy, ΔG, as:

ΔG = ΔH - TΔS

Where ΔH is enthalpy, T absolute temperature and ΔS change in entropy.

Replacing (25°C = 273 + 25 = 298K; 25.45kJ/mol = 25450J/mol):

ΔG = ΔH - TΔS

ΔG = 25450J/mol - 298K×108.7J/molK

ΔG = -6942.6J/mol

Right solution is:

<h3>B)−6,942 J /mol</h3>

8 0
3 years ago
The charge of single electron
irga5000 [103]
The answer is B.

plz mark me as brainliest. i really need it.
8 0
3 years ago
The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the c
Alenkinab [10]

<u>Answer:</u> The spacing between the crystal planes is 4.07\times 10^{-10}m

<u>Explanation:</u>

To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

n\lambda =2d\sin \theta

where,

n = order of diffraction = 2

\lambda = wavelength of the light = 154pm=1.54\times 10^{-10}m     (Conversion factor:  1m=10^{12}pm )

d = spacing between the crystal planes = ?

\theta = angle of diffraction = 22.20°

Putting values in above equation, we get:

2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m

Hence, the spacing between the crystal planes is 4.07\times 10^{-10}m

4 0
3 years ago
Comment Both propane and benzene are hydrocarbons. As a rule,
kozerog [31]

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

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2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

5 0
3 years ago
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