Answer:
A particle
Explanation:
Modern quantum theory holds that light has both wave-like and particle-like properties. When the length scales involved are large compared to the wavelengths of light (ex., forming images with thin lenses), the
particle nature of light dominates.
Answer: Nuclei combine to form a heavier nucleus, releasing energy.
Explanation: e.g two deuterium nucleus (Hydrogen-2 isotopes) forms an He nucleus and energy is released.
Answer: The wavelength of the x-ray wave is 
Explanation:
To calculate the wavelength of light, we use the equation:

where,
= wavelength of the light = ?
c = speed of x-ray= 
= frequency of x-ray =

Putting in the values:

Thus the wavelength of the x-ray wave is 
a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>
Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

volume NO at 1273 K and 1 atm

b. 15 L NH3 at STP ( 1mol = 22.4 L)

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

mass H2O(MW = 18 g/mol) :

c. mol NO at 1273 K and 1 atm :

mol ratio of NO : O2 = 4 : 5, so mol O2 :

Volume O2 at STP :

I believe it can be warm because usually the surface below us is warm itself, causing the air to be warm as well. The temperature can vary as well