Respiration and Photosynthesis is different form each other because photosynthesis removes CO2 while respiration puts back CO2
I hoped this helped!
No a flame doesn't always give heat because sometimes fire can burn but it doesn't have heat
Answer:
a) Acidic buffer
b) No buffer
c) Acidic buffer
d) Basic buffer
e) Basic buffer
Explanation:
a) 75.0 mL of 0.10 M HF ; 55.0 mL of 0.15 M NaF
-Acidic buffer
Mixing of 75.0 mL of 0.10 HF and 55.0 mL of 0.15 mL NaF results in acidic buffer. HF/NaF is a buffer of weak acid and its conjugate base. F- is the conjugate base of acid,HF.
b.) 150.0 mL of 0.10 M HF ; 135.0 mL of 0.175 M HCl-No buffer
Mixing HF and HCl will not results in a buffer. Both are acids, and no conjugate base is present.
c.) 165.0 mL of 0.10 M HF ; 135.0 mL of 0.050 M KOH-Acidic buffer
HF reacts with KOH to form KF. F- is a conjujate base of HF. As volume and concentration of HF is more than KOH, therefore, HF will remain after reaction with KOH. HF/KF will be a buffer of weak acid and its conjugate base.
d.) 125.0 mL of 0.15 M CH3NH2 ; 120.0 mL of 0.25 M CH3NH3Cl -Basic buffer
CH3NH2/CH3NH3+ is a buffer of weak base and its conjugate acid.
e.) 105.0 mL of 0.15 M CH3NH2 ; 95.0 mL of 0.10 M HCl-Basic buffer
CH3NH2 is a weak base and HCl is a strong acid. CH3NH2 reacts with HCl to form its conjugate acid CH3NH3+. Volume and concentration of CH3NH2 is more as compared to HCl and hence, will remain in the soution after reactionf with HCl.
CH3NH3+/CH3NH2 is a buffer of weak base and its conjugate acid.
The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation
R = molar gas constant
K = A(e^(-Ea/RT))
Taking natural log of both sides
In K = In A - (Ea/RT)
In K = (-Ea/R)(1/T) + In A
Comparing this to the equation of a straight line; y = mx + c
y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A
a) From the question, m = (-Ea/R) = -1.10 × (10^4) K
(-Ea/R) = -1.10 × (10^4) = -11000
R = 8.314 J/K.mol
Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol
b) c = In A = 33.5
A = e^33.5 = (3.54 × (10^14))/s
c) K = A(e^(-Ea/RT))
A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol
K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s
QED!
