Answer:
Explanation:
1. It is not Blanced. on the right the is 1S and 6O and the left there is 1S and 3O
2.If the energy level of the reactants is higher than the energy level of the products the reaction is exothermic (energy has been released during the reaction). If the energy level of the products is higher than the energy level of the reactants it is an endothermic reaction.
Answer:
The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.
Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.
Explanation:
CH4(g) + 2H2O(g) ⇆ CO2(g) + 4H2(g)
CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0
ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0
S°(J/K·mol): 186.1 188.8 213.7 130.7
ΔG<0 to be spontaneous
ΔG = ΔH- TΔS <0
ΔH = ∑nΔH(products) - ∑nΔH(reactant)
ΔH = (-393.5) - (–74.87 + 2*–241.8)
ΔH = 164.97 kJ = 164970 J
ΔS = ∑nΔS(products) - ∑nΔS(reactant)
ΔS = (213.7 + 4*130.7) - (186.1 + 2*188.8)
ΔS = 172.8 J
0 > 164970 J - T* 172.8 J
-164970 J > - T* 172.8 J
954.7< T
The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.
Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.
Answer:
pH = 4.8
Explanation:
A buffer is formed by a weak acid (0.145 M HC₂H₃O₂) and its conjugate base (0.202 M C₂H₃O₂⁻ coming from 0.202 M KC₂H₃O₂). The pH of a buffer system can be calculated using Henderson-Hasselbalch's equation.
![pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%5C%5CpH%20%3D%20-log%281.8%20%5Ctimes%2010%5E%7B-5%7D%20%29%2Blog%28%5Cfrac%7B0.202M%7D%7B0.145M%7D%20%29%5C%5CpH%3D4.8)
Answer:
0.0055 mol of N2O5 will remay after 7 min.
Explanation:
The reaction follows a first-order.
Let the concentration of N2O5 after 7 min be y
Rate = Ky = change in concentration of N2O5/time
K is rate constant = 6.82×10^-3 s^-1
Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M
Change in concentration = 0.0117 - y
Time = 7 min = 7×60 = 420 s
6.82×10^-3y = 0.0117 - y/420
0.0117 - y = 420×6.82×10^-3y
0.0117 - y = 2.8644y
0.0117 = 2.8644y + y
0.0117 = 3.8644y
y = 0.0117/3.8644 = 0.00303 M
Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)
Answer:
C) exothermic
Explanation:
The given reaction is exothermic.
N₂ + 3H₂ → 2NH₃ + ENERGY
when energy is released the reaction is exothermic and when energy is written on left side with reactant it means energy is added and reaction is endothermic.
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
