Earth orbits the sun in a imaginary plane called ecliptic
The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>
Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :

NO :

Cl2 :

![\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BNO%5D%5E2%5BCl_2%5D%7D%7B%5BNOCl%5D%5E2%7D%5C%5C%5C%5CKc%3D%5Cdfrac%7B0.224%5E2%5Ctimes%200.112%7D%7B0.464%5E2%7D%3D0.026)
Molar mass neon = <span>20.1797 g/mol
20.1797 g ---------- 6.02x10</span>²³ atoms
? g ------------ 3.0 x 10²³ atoms
mass = 3.0 x 10²³ * <span>20.1797 / 6.02 x 10</span>²³
mass = 6.054x10²⁴ / 6.02 x 10²³
mass = 10.05 g
hope this helps!
The (15-35-15) number on a bag of fertilizer is calculated by an outdated method. Using the knowledge of the old methods, the number of moles of phosphorus found in 10.0 grams of phosphorus is 0.05 mole.
The (15-35-15) number on the bag of fertilizer is used for:
- Nitrogen 15 :
Phosphorus pentoxide 35: potassium oxide 15
Thus, in a 100-gram bag of fertilizer, there are 35 grams of phosphorus pentoxide P₂O₅.
∴
To determine the number of moles of 10 grams of phosphorus pentoxide in (15-35-15) bag of fertilizer, we have:

= 3.5 grams of P₂O₅
Recall that:

- the molar mass of P₂O₅ = 142 g/mol
∴


Provided that 1 mole of P₂O₅ harbors 2 moles of phosphorus;
Then;
The number of phosphorus present in 10.0 grams of (15-35-15) fertilizer is:
= 0.0246 moles × 2
= 0.0492 moles
≅ 0.05 moles
Therefore, we can conclude that the number of moles of phosphorus found in 10.0 grams of (15-35-15) bag of fertilizer is 0.05 moles.
Learn more about phosphorus here:
brainly.com/question/4622631?referrer=searchResults
Answer:
a) and d) are false.
Explanation:
a) The second law of thermodynamics states that t<u>he entropy of the universe increases in a spon
taneous process and remains unchanged in an equilibrium process.</u>
For a spontaneous process:
ΔSuniv = ΔSsys + ΔSsurr > 0
For a spontaneous process, the second law says that ΔSuniv must be greater than zero, but it does not place a restriction on either ΔSsys or ΔSsurr. Thus, it is possible for either ΔSsys or ΔSsurr to be negative, as long as the sum of these two quantities is greater than zero.
d) <u>A reaction that does occur under the given set of conditions is called a</u> spontaneous reaction. We observe spontaneous physical and chemical processes every day, including many of the following examples:
• A waterfall runs downhill, but never up, spontaneously.
• A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar does not spontaneously reappear in its original form.
• Heat flows from a hotter object to a colder one, but the reverse never happens spontaneously.