Answer:
Group 12
Explanation:
Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.
Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.
Answer:
D. because water and mud behaved in a similar way in the past as they do today
Explanation:
One of the fundamental theories in the field of earth science is the theory of uniformitarianism.
Uniformitarianism was proposed by James Hutton in the 18th century in Scotland. The theory states that "geologic processes occurring today have occurred in times past and that the present is the key to past".
The simple meaning of the theory is that, the processes on earth today such as weathering, erosion, e.t.c have also occurred in times past. Those processes still occur today and an understanding of such events today will help us have a better insight into the past.
Therefore, ripple marks just as they form today from action of mud and water would be formed in a similar way in the past.
233.856 , sorry if i’m wrong
Answer:
About one valence electron
Explanation:
Obviously, removing that electron gives us [Ar] (same configuration as K1+), which is a noble gas and has 8 electrons. Valence electrons are generally regarded as being 'the outermost electrons' for a given atom. Therefore, with neutral potassium, there is one valence electron
Answer:
1.427x10^-3mol per L
Explanation:
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L
