The answer is A: When the energy transporting sediments diminishes, the sediments settle in a low-lying area; therefore, deposition always follows erosion
It is B, and also for a moment I didn't understand that 4.69 x 10^22. I almost did this whole problem wrong.
Answer:
it identifies reducing sugars (monosaccharide's and some disaccharides), which have free ketone or aldehyde functional group
Explanation:
It turns from turquoise to yellow or orange when it reacts with reducing sugars.
The lighter components are able to rise higher in the column before they are cooled to their condensing temperature, allowing them to be removed at slightly higher levels.
I hope this helps
We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.
