The ability for gas to mix with other substances is diffusion.
C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole
D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles
<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>
1 mole of N₂H₄ contains 4 moles of H
Therefore,
0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H
<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>
1 mole of C₁₀H₂₂ contains 22 moles of H
Therefore,
34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H
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Answer:
Waves involve the transport of energy without the transport of matter. In conclusion, a wave can be described as a disturbance that travels through a medium, transporting energy from one location (its source) to another location without transporting matter.
Explanation:
hope this helps, I tried to give an easy definition :)
The second one, I could be mistaken though
194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have
moles.
So 
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.


So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.