Answer:
0.00915 M of 
remain after 5.16 seconds.
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
s⁻¹
Initial concentration = 0.054 M
Final concentration = ? M
Time = 5.16 s
Applying in the above equation, we get that:-
![[A_t]=0.054e^{-0.344\times 5.16}\ M=\frac{1\times \:0.054}{e^{1.77504}}\ M=0.00915\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.054e%5E%7B-0.344%5Ctimes%205.16%7D%5C%20M%3D%5Cfrac%7B1%5Ctimes%20%5C%3A0.054%7D%7Be%5E%7B1.77504%7D%7D%5C%20M%3D0.00915%5C%20M)
<u>0.00915 M of remain after 5.16 seconds.</u>
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)
Molality= mol/ Kg
if we assume that we have 1 kg of water, we have 3.19 moles of solute.
the formula for mole fraction --> mole fraction= mol of solule/ mol of solution
1) if we have 1 kg of water which is same as 1000 grams of water.
2) we need to convert grams to moles using the molar mass of water
molar mass of H₂O= (2 x 1.01) + 16.0 = 18.02 g/mol
1000 g (1 mol/ 18.02 grams)= 55.5 mol
3) mole of solution= 55.5 moles + 3.19 moles= 58.7 moles of solution
4) mole fraction= 3.19 / 58.7= 0.0543
