Answer:
We might just have to end it together
Explanation:
I tried to answer it now I'm stuck in the same hole -_-
a) To find the mass after t years:we will use this formula:
A = Ao / 2^n when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)b) Mass after 90 y :by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mgC) Time for 1 mg remaining:when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
Answer: a. 0.26mol
b. 0.000479mol
c. 1.12mol
Explanation: Please see attachment for explanation
Answer:
The correct statements are:
The rate of disappearance of B is twice the rate of appearance of C.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.
3A + 2B → C + 2D
Rate of the reaction:
![R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
The rate of disappearance of B is twice the rate of appearance of C.
![\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules