Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Hey there:
Correct answer is :
(b) NaNH₂
Sodium azanide NaNH₂ is the conjugate base of ammonia NH₃
Correct answer is :
(b) NaNH₂
I hope this will help !
Answer:
![\large \boxed{\text{B.) 2.8 atm}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7BB.%29%202.8%20atm%7D%7D)
Explanation:
The volume and amount are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
![\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bp_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bp_%7B2%7D%7D%7BT_%7B2%7D%7D)
Data:
p₁ = 1520 Torr; T₁ = 27 °C
p₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 27 + 273.15) K = 300.15 K
T₂ = (150 + 273.15) K = 423.15 K
(b) Calculate the new pressure
![\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B1520%7D%7B300.15%7D%20%26%20%3D%20%26%20%5Cdfrac%7Bp_%7B2%7D%7D%7B423.15%7D%5C%5C%5C%5C5.064%20%26%20%3D%20%26%20%5Cdfrac%7Bp_%7B2%7D%7D%7B423.15%7D%5C%5C%5C%5C5.064%5Ctimes423.15%26%3D%26p_%7B2%7D%5C%5Cp_%7B2%7D%20%26%20%3D%20%26%20%5Ctext%7B2143%20Torr%7D%5Cend%7Barray%7D%5C%5C)
(c) Convert the pressure to atmospheres
![p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}](https://tex.z-dn.net/?f=p%20%3D%20%5Ctext%7B2143%20Torr%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20atm%7D%7D%7B%5Ctext%7B760%20Torr%7D%7D%20%3D%20%5Ctextbf%7B2.8%20atm%7D%5C%5C%5C%5C%5Ctext%7BThe%20new%20pressure%20reading%20will%20be%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B2.8%20atm%7D%7D%24%7D)
Well because the lowest possibility of having a decen is 3-decene so 6-decene is not possible. Hope this helps!! :)
Answer:
Magnesium dichloride
HoPe ThIs HeLpS! aNd HaVe A gOoD dAy!