<span>The correct way to write 602,200,000,000,000,000,000,000 in scientific notation is 6.202 x 10^23. In the scientific notation, the number is written as a multiplication of a number from 1 to 9 and 10 raised to the adequate power. 602,200,000,000,000,000,000,000 = 6.202 x 100,000,000,000,000,000,000,000. Since 10 = 10^1; 100 = 10^2; 1,000 = 10^3, etc. then 100,000,000,000,000,000,000,000 = 10^23. Therefore, 602,200,000,000,000,000,000,000 = 6.202 x 100,000,000,000,000,000,000,000 = 6.202 x 10^23.</span>
Scientific notation : It is defined as the representation of expressing the number that are too big or too small that is written in the decimal form. That means always written in the power of 10 form.
For example : 5000 is written as,
As we are given that the value is, 602,200,000,000,000,000,000,000
This number is written in scientific notation as :
Therefore, the correct way to write the given number is,
a) The magnitude of intermolecular forces in compounds affects the boiling points of the compound. Neon has London dispersion forces as the only intermolecular forces operating in the substance while HF has dipole dipole interaction and strong hydrogen bonds operating in the molecule hence HF exhibits a much higher boiling point than Ne though they have similar molecular masses.
b) The boiling points of the halogen halides are much higher than that of the noble gases because the halogen halides have much higher molecular masses and stronger intermolecular forces between molecules compared to the noble gases.
Also, the change in boiling point of the hydrogen halides is much more marked(decreases rapidly) due to decrease in the magnitude of hydrogen bonding from HF to HI. The boiling point of the noble gases increases rapidly down the group as the molecular mass of the gases increases.
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
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.