Answer:
The first option
Explanation:
Carbon is a p-block element. It is the 6th element on the periodic table and therefore it has 6 electrons.
The sub-level notation is given as:
1s² 2s² 2p²
The s-sublevel can only accommodate two maximum electrons because it has one orbital. This is why both 1s and 2s contains just two electrons each. When both sub-levels are filled, we have just 2 remaining electrons to fill the p-sublevel.
The p-sublevel contains 3 orbitals and can accommodate a maximum of 6 electrons. But we have just 2 electrons. According to Hund's rule of maximum mulitiplicity, electrons will go into degenerate orbitals singly before paring up. Therefore, the first two orbitals in p-sublevel will receive an electron each.
This is why the first model fits.
The correct answer is A. It is uniform in composition and the parts that make up the mixture can be separated from one another through physical means.
Explanation:
In a homogeneous mixture, components are completely integrated, which means the final substance is uniform and the parts that compose it are not separated. This occurs in milk because this integrates uniformly water, fat, among others, and these elements cannot be observed separately.
Moreover, in mixtures, components can be separated through physical means; for example by heating the substances. This applies to milk because if it is heated water evaporates, and therefore can be separated.
Molar mass KCl = <span>74.5513 g/mol
Number of moles:
21.9 / 74.5513 => 0.293 moles
Volume = 869 mL / 1000 => 0.869 L
Molarity = moles / Volume
Molarity = 0.293 / 0.869
=> 0.337 M</span>
Answer:
1.135 M.
Explanation:
- For the reaction: <em>2HI → H₂ + I₂,</em>
The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².
- To solve this problem, we can use the integral law of second-order reactions:
<em>1/[A] = kt + 1/[A₀],</em>
where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),
t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),
[A₀] is the initial concentration of HI ([A₀] = ?? M).
[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).
∵ 1/[A] = kt + 1/[A₀],
∴ 1/[A₀] = 1/[A] - kt
∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.
∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.
<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>
