Alkali metals are rarely found in pure form because they are so reactive.
Alkali metals are in Group 1 of the Periodic Table.
They need to lose only one valence electron to reach a filled valence shell
Thus, they readily react wirh water and other chemicals in the earth to form compounds.
Answer:
Is this a Q? i would like to help but what the heck is this?
Explanation:
Iodine Strontium Silver...... ..I think.
Answer:
0.200 M NaOH: 13.3 mL of the concentrated solution and complete to 20.0mL
0.150 M NaOH: 10.0 mL of the concentrated solution and complete to 20.0mL
0.100 M NaOH: 6.67 mL of the concentrated solution and complete to 20.0mL
0.050 M NaOH: 3.33 mL of the concentrated solution and complete to 20.0mL
0.025 M NaOH: 1.67 mL of the concentrated solution and complete to 20.0mL
Explanation:
It is possible to prepare a solution from a more concentrated one. In the problem, the concentrated solution is 0.300M NaOH. Thus, to prepare 20.0mL of each of the solutions you will need:
0.200 M NaOH: 20.0mL × (0.200M / 0.300M) = <em>13.3 mL of the concentrated solution and complete to 20.0mL</em>
<em>The ratio between the concentrated solution and the solution you want to prepare is called "dilution factor"</em>
<em />
0.150 M NaOH: 20.0mL × (0.150M / 0.300M) = <em>10.0 mL of the concentrated solution and complete to 20.0mL</em>
<em />
0.100 M NaOH: 20.0mL × (0.100M / 0.300M) = <em>6.67 mL of the concentrated solution and complete to 20.0mL</em>
<em />
0.050 M NaOH: 20.0mL × (0.050M / 0.300M) = <em>3.33 mL of the concentrated solution and complete to 20.0mL</em>
<em />
0.025 M NaOH: 20.0mL × (0.025M / 0.300M) = <em>1.67 mL of the concentrated solution and complete to 20.0mL</em>
The answer is the law of Conservation of Matter!
