Answer:
The correct statement is:
E - The entropy of the products is greater than the entropy of the reactants.
Explanation:
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
As glucose is a large molecule and then it is transformed into many molecules of water and carbon dioxide, the entropy of the system increases. If the number of molecules increases, the disorder increases.
Initial state: 7 molecules (1 glucose + 6 oxygen)
Final state: 12 molecules (6 carbon dioxide + 6 water)
The simple machine used is called atwood machine.
<span>9.40x10^19 molecules.
The balanced equation for ammonia is:
N2 + 3H2 ==> 2NH3
So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol
Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol
Now to convert from moles to molecules, just multiply by Avogadro's number:
1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
