The next test would be the lime water test.
Explanation:
If placing the gas in lime water will turn the solution milky, then the gas is most probably carbon dioxide, the lime water (Ca(OH)₂) turns milky due to the formation of solid CaCO₃ that is insoluble after reaction with carbon dioxide.
Carbon dioxide also puts out the flame in the splint test because it does not support combustion.
Answer:
See explanation
Explanation:
The electron configuration of an atom in an element determines the property of the atom. The core electrons are found inside the atom while the valence electrons are found on the outermost shell of the atom.
For cobalt, the outermost shell electron configuration is; [Ar] 3d7 4s2. The 3d7 and 4s2 are found in the valence shell of cobalt.
For arsenic, the electronic configuration is [Ar] 3d¹⁰ 4s² 4p³. The valence electrons are 4s2, 4p3. The 3d electrons are found inside the arsenic atom.
Pv=nRT
where,p=199, R(constant)=8.314, V=4.67 T=30C=293K
n=pv/RT=0.38 moles
Polar Covalent is the name used to describe bonds that have both ionic and covalent character because the electrons are shared unequally. The Pauling Scale is used to assign electronegativity to atoms. It ranges from to 4.00 (fluorine).
Answer: The distance is slightly less than 3.5 m
Explanation: assuming wall and target are the same thing, and the bullet has constant velocity, the bullet will travel 7 m in half a second, so half that distance is 3.5 m.
In reality, the bullet is decelerating (at an unknown rate) so the distance is slightly less than 3.5 m.
There is also a vertical velocity component, which means it hits the target/wall at an angle. The trajectory is such that it hits the wall above the shooter because the ricochet hits at ~the level at which it left the firearm.
If the wall was absent, the bullet would have described a parabola which brough it back to the initial level after 7m. This could be calculated, but it means that the actual distance between the shooter and the wall is slightly less than 3.5 m
In addition, the collision with the wall is not 100% elastic, so the velocity aftercthe ricochetvis further reduced.
A calculation would be complex because these confounding factors are not completely independent of each other, but all reduce the average velocity and therefore the distance.
Therefore it is only possible to say that the distance was somewhat less than 3.5 m
