Answer:
C I think from what I did to solve!!
52.0 divided by 65 g/mol gives you 0.8mol
Explanation:
According to mass action,
![\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2](https://tex.z-dn.net/?f=%5Ctextrm%7Brate%7D%3D-%5Cdfrac%7B%5CDelta%5B%5Ctextrm%20A%5D%7D%7B2%5CDelta%20t%7D%3Dk%5B%5Ctextrm%20A%5D%5E2)
Where, k is the rate constant
So,
![\dfrac{d[A]}{dt}=-k[A]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-k%5BA%5D%5E2)
Integrating and applying limits,
![\int_{[A_t]}^{[A_0]}\frac{d[A]}{[A]^2}=-\int_{0}^{t}kdt](https://tex.z-dn.net/?f=%5Cint_%7B%5BA_t%5D%7D%5E%7B%5BA_0%5D%7D%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%5E2%7D%3D-%5Cint_%7B0%7D%5E%7Bt%7Dkdt)
we get:
![\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20%5Cdfrac%7B1%7D%7B%5BA%5D_0%7D%2Bkt)
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
Half life is the time when the concentration reduced to half.
So, ![[A_t]=\frac{1}{2}\times [A_0]](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5BA_0%5D)
Applying in the equation as:
![t_{1/2}=\dfrac{1}{k[A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA_o%5D%7D)
Answer:
V H2O = 170.270 mL
Explanation:
- QH2O ( heat gained) = Qcoffe ( heat ceded)
⇒ Q = m<em>C</em>ΔT
∴ m: mass (g)
∴ <em>C</em>:<em> </em>specific heat
assuming:
- δ H2O = δ Coffe = 1.00 g/mL
- <em>C</em> H2O = <em>C</em> coffe = 4.186 J/°C.g....from literature
⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)
∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe
⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J
⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)
⇒ (26371.8 J)/(154.882 J/g) = m H2O
⇒ m H2O = 170.270 g
⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL
