Wilhelm Conrad Roentgen was a German scientist who discovered x-rays through the use of Crookes tube, a tube he used in studying cathode rays that emitted new kinds of invisible ray that was capable of penetrating through a black paper.
After hearing such discovery, Henri Becquerel, a French scientist had also took interest in the connection between the phosphoresence he had already been investigating and the newly discovered rays. He thought that the phosphoresence uranium salts he had been observing and studying might absorb sunlight and emit it as x-rays.
To test such idea which was disproved later on, Becquerel wrapped photographic plates in black paper so that sunlight could not reach them. He then placed the crystals of uranium on top of the wrapped plates and put the whole set up outside, exposed under the sun.
When he developed the plates, he saw an outline of the uranium crystals. He also placed other objects such as coins or cut out metals between the crystals and the photographic film/plate. It also turned out that he could also produced outlines of those shapes.
Answer =B) folding of rock
Answer:

Explanation:

Assume 1 kg water.
1. Moles of P.HCl
Then we have 4.666 mol of P.HCl
2. Mass of P.HCl
n = 4.666 mol × 272.77 g/mol = 1271.1 g
3. Total mass of solution
m = 1000 g + 1271.1 g = 2271.1 g solution
4. Volume of solution

5. Molar concentration


The balanced chemical reaction is written as:
<span>Zn + 2AgNO3 = Zn(NO3)2 + 2Ag
To determine the grams of silver metal that is being produced, it is important to first determine which is the limiting reactant and the excess reactant from the given initial amounts. We do as follows:
4.35 g Zn ( 1 mol / 65.38 g ) ( 2 mol AgNO3 / 1 mol Zn ) = 0.1331 mol AgNO3 needed
35.8 g AgNO3 ( 1 mol / 169.87 g ) ( 1 mol Zn / 2 mol AgNO3 ) = 0.1054 mol Zn needed
Therefore, the limiting reactant would be the zinc metal since it would be consumed completely in the reaction. The excess amount of AgNO3 would be:
0.2107 mol AgNO3 - 0.1331 mol AgNO3 = 0.0776 mol AgNO3 left ( 169.87 g / 1 mol ) = 13.19 g AgNO3 left
0.0665 mol Zn ( 2 mol Ag / 1 mol Zn) ( 107.9 g / 1 mol) = 14.3581 g Ag produced</span>