Answer:
1) No shift
2) No shift
3) Leftward shift
4)Rightward sifht
Explanation:
1) 2) Adding N or Removing N in the equilibrium will produce No shift, because of its solid state, the N is not contemplated in the equilibrium equation:
![K=\frac{[L]^2}{[M]^3}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BL%5D%5E2%7D%7B%5BM%5D%5E3%7D)
3) Increasing the volume produces a decrase in the preassure due to the expansion of the gases. This will cause a leftward shift, because the system will try to increase the moles of gas and in consecuence of this, also increase the preassure.
4) Decreasing the volume has the opposite effect of the item 3): the preassure will increase and the system will consume moles of gas to decrease it, producing a rightward shift.
Answer:
The correct answer is - D C2H4.
Explanation:
Saturated hydrocarbons are hydrocarbons with single covalent C-C bonds. They are known as alkanes. The general formula for these hydrocarbons is CnH2n+2
Unsaturated hydrocarbons the hydrocarbons with double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is CnH2n and CnHn-2
For the given options:
Option D: C2H4, is the simplest alkene with a double bond so it is an unsaturated hydrocarbon.
Answer:
1.81 x 10²⁴ atoms
Explanation:
To find the number of atoms in the given number of moles, we need to understand that every substance contains the Avogadro's number of particles.
More appropriately, a mole of any substance will contain the Avogadro's number of particles which is 6.02 x 10²³ atoms
So;
If 1 mole of a substance = 6.02 x 10²³ atoms;
3 mole of MgCl₂ will contain 3 x 6.02 x 10²³ = 1.81 x 10²⁴ atoms
The activity of the sample when it was shipped from the manufacturer is 4.54 mCi
<h3>How to determine the number of half-lives that has elapsed </h3>
From the question given above, the following data were obtained:
- Time (t) = 48 hours
- Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
- Number of half-lives (n) =?
n = t / t½
n = 48 / 342.72
n = 0.14
<h3>How to determine the activity of the sample during shipping </h3>
- Number of half-lives (n) = 0.14
- Original activity (N₀) = 5.0 mCi
- Activity remaining (N) =?
N = N₀ / 2ⁿ
N = 5 / 2^0.14
N = 4.54 mCi
Thus, the activity of the sample during shipping is 4.54 mCi
Learn more about half life:
brainly.com/question/2674699
