Equation is as follow,
<span> CaCN</span>₂<span> + 3 H</span>₂<span>O → CaCO</span>₂<span> + 2 NH</span>₃
According to this equation,
When,
100 g CaCO₃ (1 mole) is produced when = 34 g NH₃ (2 moles) is produced
So,
187 g CaCO₃ will be produced then = X g of NH₃ will produce
Solving for X,
X = (187 g × 34 g) ÷ 100 g
X = 63.58 g of NH₃ will be produced
Result:
Option-2 is correct answer.
Answer:
Generally SN1 reactions are carried out in polar protic solvents. The energy required for breaking the C-X bond is obtained through solvation of halide ion with the proton of the solvent. Tertiary alkyl halides undergo SN1 reaction very fast because of the high stability of tertiary carbocations.
Answer:
You must write an article if you think there is a link between it asteroid impact and a dinosaur's extinction. explain the correlation between the two events I'm provide evidence within the article that you write that supports your claims.
Answer:
Explanation:
By definition one <em>half-life</em> is the time to reduce the initial concentration to half.
For a <em>second order reaction </em>the rate law equations are:
![\dfrac{d[B]}{dt}=-k[B]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D-k%5BB%5D%5E2)
![\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BB%5D%7D%3D%5Cdfrac%7B1%7D%7B%5BB%5D_0%7D%2Bkt)
The <em>half-life</em> equation is:
![t_{1/2}=\dfrac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%7D)
Thus, substitute the<em> rate constant</em> 
and the <em>half-life </em>time <em>224s</em> to find [A]₀:
![224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}](https://tex.z-dn.net/?f=224s%3D%5Cdfrac%7B1%7D%7B1.30%5Ctimes10%5E%7B-3%7DM%5E%7B-1%7D%5Ccdot%20s%5E%7B-1%7D%5BA%5D_0%7D)
![[A]_o=0.291M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.291M)
