Answer:
Explanation:
Homogeneous mixture is a mixture in which the components of the mixture are in the same proportion throughout any sample extracted from the mixture while an heterogeneous mixture is a mixture in which the components of the mixture differ in term of proportion when different samples of the mixture are extracted and compared.
For example, a sandy water will have some parts (usually the bottom) of the mixture with more sand than other parts of the mixture, hence, it (sandy water) is a heterogeneous mixture. While salty and ocean water has it's salt dissolved in the same proportion throughout the water in the mixture, hence salty and/or ocean water is a homogeneous mixture.
Sandy water can be separated by filtration (i.e using a filter paper to separate the sand from the water when the mixture is poured over a filter paper) while salty and ocean water can be separated by distillation (i.e boiling of the mixture so the water molecules can boil and move through a tube as gas or steam into another container where they are cooled and converted back to liquid or water while leaving the solid salt component of the mixture in the boiling tube).
<span>I would say only if one of your data points is the origin. But your experiment could have started with a non-zero velocity, for instance, which would rule out the origin as one of your data points. Even so, a "best fit" is not meant to be perfect, it is only meant to be the best that you can do with your particular data set.</span>
Answer:
Correct choice are C and D (they are both, the same).
Explanation:
Cathode is the positive(+) electrode where a reduction occurs.
Reduction is the chemical reaction where the oxidation state is reduced.
2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)
A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
B. 2Ag (s) → 2Ag+ (aq) + 2e-
C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
C or D, are ok. They are the same equation.
Oxygen from ground state reduce the oxidation state from 0 to -2
Electronic configuration of cromium is
Cr-[Ar]4s¹3d⁵
When cromium loses two electrons it becomes Cr⁺².
So its electronic configuration becomes,
Cr⁺²-[Ar]3d⁴
One electron will go from 4s orbital and one electron will go from 3d orbital.
So the answer here is D. [Ar]3d⁴ -because after losing 2 electrons electronic configuration of cromium becomes [Ar] 3d⁴.
