Answer:
10.000 grams
Explanation:
For the first law of thermodynamics, the energy must be conserved, that means that the energy in form of heat (Q) must be equal to the sum of work (W) and internal energy(ΔU) :
Q = W + ΔU
ΔU depends on the temperature and W in the variation of pressure and volume. Q depends on the temperature, but also the mass. So, there is the same temperature, ΔU is equal for both reaction, if there is no work done, the heat must be equal for both of them. So the mass such be the same.
Answer:
I believe 11 is B 12 is C 13 is B and 14 is C
Answer:
40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[magenta(Php)]}{[Php]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bmagenta%28Php%29%5D%7D%7B%5BPhp%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of phenolphthalein = 9.40
= concentration of magenta phenolphthalein
= concentration of colorless phenolphthalein
pH = 11
Putting values in above equation, we get:
![11=9.40+\log(\frac{[magenta(Php)]}{[Php]})](https://tex.z-dn.net/?f=11%3D9.40%2B%5Clog%28%5Cfrac%7B%5Bmagenta%28Php%29%5D%7D%7B%5BPhp%5D%7D%29)
![\log(\frac{[magenta(Php)]}{[Php]})=11-9.40=1.6](https://tex.z-dn.net/?f=%5Clog%28%5Cfrac%7B%5Bmagenta%28Php%29%5D%7D%7B%5BPhp%5D%7D%29%3D11-9.40%3D1.6)
![\frac{[magenta(Php)]}{[Php]}=10^{1.6}=39.81 :1 \approx 40:1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bmagenta%28Php%29%5D%7D%7B%5BPhp%5D%7D%3D10%5E%7B1.6%7D%3D39.81%20%3A1%20%5Capprox%2040%3A1)
40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.
Answer:
E_a = 103.626 × 10³ KJ/mol
Explanation:
Formula to solve this is given by;
Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))
Where;
k2 is rate constant at second temperature
k1 is rate constant at first temperature
R is universal gas constant
T1 is first temperature
T2 is second temperature
We are given;
k1 = 2.8 × 10^(-3) /s
k2 = 4.8 × 10^(-4) /s
R = 8.314 J/mol.k
T1 = 60°C = 333.15 K
T2 = 45°C = 318.15 K
Thus;
Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))
We now have;
-0.76592 = -0.00000739121E_a
E_a = -0.76592/-0.00000739121
E_a = 103.626 × 10³ KJ/mol