Hey there!:
Molar mass N₂ = 28.0134 g/mol
28.0134 g ------------------- 22.4 L (at STP )
mass N₂ -------------------- 50.0 L
mass N₂ = 50.0 x 28.0134 / 22.4
mass N₂ = 1400.67 / 22.4
mass N₂ = 62.529 g
Hope this helps!
To covert from moles to atoms times the number of moles by Avogadro's Number (6.022×10²³)
4.0 × 6.022×10²³ = 2.4088×10^24
It seems that you have missed the given image to answer this question. But anyway, I found it and got the answer. Based on the topographical map of a section of Charleston, SC, the feature that is <span>located at the dot marked with an X is the high point of a hill. The answer would be option D.</span>
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol
1 mole=6.022*10²³ particles (atoms or molecules)
Then: 6.022*10²³ atoms are contained in 20.18g
Now, We can solve this problem by the three rule.
6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g
x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.
Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.
