The answer is (2). If you recall Rutherford's gold foil experiment, remember that a stream of positively charged alpha particles were shot at a gold foil in the center of a detector ring. The important observation was that although most of the particles passed straight through the foil without being deflected, a tiny fraction of the alpha particles were deflected off the axis of the shot, and some were even deflected almost back to the point from which they were shot. The fact that some of the alpha particles were deflected indicated a positive charge (because same charges repel), and the fact that only a small fraction of the particles were deflected indicated that the positive charge was concentrated in a small area, probably residing at the center of the atom.
Answer:
C) H2S
Explanation:
In chemistry, the dissolution of one substance in another is dependent on the magnitude of intermolecular interaction between the two substances. Hence, if two substances do not interact in one way or the other, then one can not dissolve the other.
Let us consider the fact that NH3 is a polar molecule and it is a general principle that like dissolves like. Hence, only H2S which is also a polar molecule can effectively interact with NH3 due to dipole-dipole interaction between the two molecules.
Also, ammonia reacts with hydrogen sulphide as follows;
2NH3 + H2S → (NH4)2S
Hence H2S is more likely to dissolve in NH3.
₁₇Cl 1s²2s²2p⁶3s²3p⁵, 1 unpaired electron in 3pz orbital.
pH solution = 8.89
<h3>Further explanation</h3>
Given
The concentration of HBr solution = 1.3 x 10⁻⁹ M
Required
the pH
Solution
HBr = strong acid
General formula for strong acid :
[H⁺]= a . M
a = amount of H⁺
M = molarity of solution
HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1
Input the value :
[H⁺] = 1 x 1.3 x 10⁻⁹
[H⁺] = 1.3 x 10⁻⁹
pH = - log [H⁺]
pH = 9 - log 1.3
pH = 8.89
Answer:
Explanation:
How many mols do you have?
1 mol = 6.02 * 10^23 atoms
x mol = 6.25 * 10 ^32 atoms
1/x = 6.02*10^23 / 6.25 * 10^32 Cross multiply
6.02 * 10^23 * x = 1 * 6.25 * 10^32 Divide by 6.02 * 10^23
x = 6.25 * 10*32/ 6.02 ^10^23
x = 1.038 * 10^9 mols which is quite large.
Find the number of grams. (Use the value for copper on your periodic table. I will just use an approximate number.)\
1 mol of copper = 63 grams.
1.038 * 10^9 mols of copper = x
1/1.038 * 10^9 = 63/x Cross multiply
x = 1.038 * 10^9 * 63
x = 6.54 * 10^10 grams of copper.
