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scZoUnD [109]
3 years ago
10

Which of the following radiations consists of a helium nucleus emitted by some radioactive substances? Alpha radiation Beta radi

ation Electromagnetic radiation Gamma radiation
Chemistry
1 answer:
erica [24]3 years ago
5 0

Answer: Alpha radiation

Explanation: Alpha decay : When a larger radioactive nuclei decays into smaller nuclei by releasing alpha radiation, the mass number and atomic number is reduced by 4 and 2 units respectively.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\textrm{He}


Beta decay : When a larger radioactive nuclei decays into smaller nuclei by releasing beta radiation, the atomic number is increased by 1 unit.


_Z^A\textrm{X}\rightarrow _{Z+1}^{A}Y+_{-1}^0e


Gamma decay : When a larger radioactive nuclei decays into smaller nuclei by releasing gamma radiation, the mass number remains same.

_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma

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an atom has 70 protons, 70 electrons, and 99 neutrons. what is the mass number? group of answer choices
Blizzard [7]

Answer:

169

Explanation:

protons+neutrons=atomic mass

70+99=169

6 0
3 years ago
Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

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

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

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4 years ago
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Answer:

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