Answer:
Ka = ( [H₃O⁺] . [F⁻] ) / [HF]
Explanation:
HF is a weak acid which in water, keeps this equilibrium
HF (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + F⁻ (aq) Ka
2H₂O (l) ⇄ H₃O⁺ (l) + OH⁻ (aq) Kw
HF is the weak acid
F⁻ is the conjugate stron base
Let's make the expression for K
K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]
K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]
K . [H₂O] = Ka
Ka, the acid dissociation constant, includes Kwater.
Answer:
a) First-order.
b) 0.013 min⁻¹
c) 53.3 min.
d) 0.0142M
Explanation:
Hello,
In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.
a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.
b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

c) Half life for first-order kinetics is computed by:

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

Best regards.
Answer: Final temperature of the gas will be 330 K.
Explanation:
Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.
(At constant volume and number of moles)

where,
= initial pressure of gas = 1.00 atm
= final pressure of gas = 1.13 atm
= initial temperature of gas =
K
= final temperature of gas = ?


Therefore, the final temperature of the gas will be 330 K.
Answer:
<em>Dry suction chest tube system</em>
Explanation:
<em>The dry suction drains that are self-regulating today use a small, adjustable regulator installed into the drain. </em>
So long as there is enough air flow from the suction of the wall (that will be shown on the drain) it will respond correctly to changes in the pressure of the source or the patient to retain suction at the drain level.
It's also a quiet mode of activity.
The solid melts and becomes a liquid.