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Maksim231197 [3]
2 years ago
15

In one to two sentences, explain how to compare the strength of electrostatic forces between molecules in salt water to the stre

ngth of electrostatic forces between molecules in freshwater. Include a description of the data to be collected and how that data would allow one to draw conclusions about the relative strengths of the electrostatic forces in the two substances.
Chemistry
1 answer:
Elden [556K]2 years ago
5 0

The boiling point of salt water is greater than that of fresh water hence the electrostatic forces between molecules in salt water is greater than that between the molecules in salt water.

Fresh water contains does not contain dissolved solutes like salt water. Salt water contains a lot of dissolved mineral salts making the salt water to be rich in ions.

Generally, the boiling points of substances can serve as a measure of the strength of electrostatic forces between molecules in either kind of water. We will easily find the that the boiling point of salt water is far higher than that of fresh water indicating greater electrostatic forces between molecules in salt water than between molecules in fresh water.

Learn more about boiling point: brainly.com/question/2153588

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Combining Boyle's law and Charles' law forms one presure-temperature-volume relationship and holds the mass of the gas constant.
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(P1V1/T1)=(P2V2/T2)

I just took the test and this should be correct

4 0
3 years ago
20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
kirill115 [55]

Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
  • V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L.

Calculate the number of moles of \rm H_2SO_4 formula units in that 0.02\; \rm L of the 0.09\; \rm M solution:

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}.

Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

Previous calculations show that 0.0018\; \rm mol of \rm H_2SO_4 was produced. Calculate the number of moles of \rm NaOH formula units required to neutralize that

\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}.

Calculate the concentration of a 0.03\; \rm L solution that contains exactly 0.0036\; \rm mol of \rm NaOH formula units:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}.

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How many molecules are in 8.3 moles?
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1 mole=6.02 ×10

8.3 moles=?

8.3×6.02×10=499.66 molecules

therefore ghe no. of molecules = 499.66
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