The red and blue will spread from the drops which will mix causing purple at the meeting and red pat that then over time it will spread to a murky purple
Answer:
decreases
Explanation:
the more conc of H ions increases the more it becomes acidic so the pH goes down
Answer:
<span>Carbon readily forms covalent bonds with other carbon atoms.
Explanation:
As we know approximately more than 95 % compounds, either isolated, discovered or synthesized belongs to organic compounds containing carbon atoms.
This great diversity of organic compounds is due to following facts.
1) Catenation:
Carbon has a peculiar behavior of self linkage. This self linkage of one carbon with another is called as catenation. In this way carbon can form a long chain of carbon atom. A branching can also take place when one carbon is bonded further to three of four carbon atoms.
2) Isomerism:
Secondly the carbon containing compounds show isomerism. In which molecular formula is same but structural formula is different. For example molecular formula C</span>₅H₁₂ can make following compounds,
a) n-Pentane
b) 2-Methylbutane
c) 2,2-Dimethylpropane
3) Multiple Bonds:
Carbon can form multiple bonds i.e double bond like in alkenes and triple bonds like in alkyne.
Due to these factors carbon gets very high number of opportunities to form large number of compounds.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
