A beachside all objects have thermal energy but thermal energy is the sum of the energy of all the particles so the more particles the more energy.
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
I believe it would be a compound.
Celsius: -11.7
Kelvin: 261.5
Hope it helps! Please mark Brainliest! :)
<h3>
Answer:</h3>
10.6 mol NO
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O
[Given] 13.2 mol O₂
<u>Step 2: Identify Conversions</u>
[RxN] 5 mol O₂ → 4 mol NO
<u>Step 3: Stoich</u>
- [DA] Set up:
![\displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})](https://tex.z-dn.net/?f=%5Cdisplaystyle%2013.2%20%5C%20mol%20%5C%20O_2%28%5Cfrac%7B4%20%5C%20mol%20%5C%20NO%7D%7B5%20%5C%20mol%20%5C%20O_2%7D%29)
- [DA] Multiply/Divide [Cancel out units]:
![\displaystyle 10.56 \ mol \ NO](https://tex.z-dn.net/?f=%5Cdisplaystyle%2010.56%20%5C%20mol%20%5C%20NO)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
10.56 mol NO ≈ 10.6 mol NO